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Unformatted text preview: Calculations In Chemistry
Modules 19 and above have been renumbered.
The former Module 29 on BrønstedLowry Definitions is now Module 31
If you are looking for BrønstedLowry Definition topics, check Module 31
At www.ChemReview.Net ***** Module 29: AcidBase Fundamentals
Module 29 – AcidBase Fundamentals....................................................................... 910
Lesson 29A:
Lesson 29B:
Lesson 29C:
Lesson 29D:
Lesson 29E:
Lesson 29F: AcidBase Math Review ................................................................................... 910
Kw Calculations: H+ and OH─....................................................................... 913
Strong Acid Solutions ....................................................................................... 917
The [OH─] in Strong Acid Solutions .............................................................. 922
Strong Base Solutions........................................................................................ 925
The pH System................................................................................................... 928 For additional modules, visit www.ChemReview.Net ©2007 ChemReview.net v. 2a Page i Module 29 — AcidBase Fundamentals Module 29 — AcidBase Fundamentals
Timing: The naming of acids was covered in Lesson 7D. Acidbase neutralization
calculations were covered in Module 14. Begin this module when you are assigned
calculations that include Kw or pH.
***** Lesson 29A: AcidBase Math Review
Pretest: This lesson reviews the math frequently used in acidbase calculations. If your
exponential math is good, simply do the last lettered part of the Practice questions 58.
If more review is needed, study the rules below, then do more of the practice set. If more
detailed review is needed, see Lessons 1A to 1C.
***** Rules For Exponential Notation
1. To multiply exponential terms, add the exponents.
2. To divide exponentials, subtract the exponents. sign ↓
— 5.25 x 103
^
^
significand
exponential 3. Numbers expressed in exponential notation have three parts: 4. Answers in exponential notation should be converted to scientific notation. This places
the decimal in the significand after the first digit that is not a zero.
• Moving the decimal point Y times to make the significand larger, make the exponent
of 10 smaller by a count of Y. • Moving the decimal point Y times to make the significand smaller, make the
exponent larger by a count of Y. “If you make one larger, make the other smaller.”
5. When changing fixed decimal numbers to exponential notation, the number of times you
move the decimal becomes the positive or negative number in the power of the
exponent.
6. In scientific notation, numeric values larger than one will have positive significands and
positive powers of 10. Values between zero and one will have positive significands and
negative powers of 10.
7. In calculations using exponential notation, handle the two parts separately: do number
math by number rules and exponential math by exponential rules.
8. If an exponential term does not have a number in front, add a “1 x ” in front of the
exponential so that the numbernumber division is clear. © 2011 ChemReview.Net v. 2m Page 910 Module 29 — AcidBase Fundamentals Practice
Do the problems below without looking at the rules. If you find that you need to look back,
write a summary of the rules above, recite your rules, then continue with the practice.
To speed your progress, do every second or third problem. If you cannot solve easily, do
more parts of that problem. Check your answers at the end of this lesson.
Try problems 17 without a calculator.
If you need help, see Rule 1. a. (10─8)(10+2) = ___________ b. (10─3)(10─12) = 1. c. (x)(10─12) = 10─14 ; x = _______ ___________ d. [H+](10─9) = 10─14 ; [H+] = _______ e. (10─3) [OH─] = 10─14 ; [OH─] = _______ 2. Need a hint on the following? See Rule 2.
a. 10─14 = b. 103 10─14 =
10─5 10─14 =
10─11 c. 3. Convert these to scientific notation. For help, see Rule 3.
a. 324 x 10+12 = ___________ b. 0.050 x 10─11 = c. 0.65 x 10─7 = ___________ d. 879 x 10─5 = ___________ ___________ 4. Convert these numbers to scientific notation. See Rule 4 for help.
a. 5,260 = b. 0.010 = c. 0.0075 = 5. Write the decimal equivalents of these fractions. Practice so you can answer these from
automatically from memory.
a. 1/2 = b. 1/20 = c. 1/3 = d. 1/4 = e. 1/5 = f. 1/50 = g. 1/8 = h. 1/80 = 6. Convert your final answer to scientific notation. Need a hint? See Rule 6.
a. (2.0 x 101) (3.0 x 10─11) =
b. 1.0 x 10─14 = c. 2.0 x 104 1.0 x 10─14
3.0 x 10─4 = d. [H+](2.0 x 10─8) = 10 x 10─15 ; [H+] =
e. (2.5 x 10─2) [OH─] = 10 x 10─15 ; © 2011 ChemReview.Net v. 2m [OH─] = Page 911 Module 29 — AcidBase Fundamentals 7. Try to solve without a calculator. It may help to convert all values to scientific notation.
a. 1.0 x 10─14 =
0.040 b. 1.0 x 10─14
0.0030 = c. [H+](0.20) = 1.0 x 10─14 ; [H+] =
d. (0.0125) [OH─] = 1.0 x 10─14 ; [OH─] = 8. Use a calculator for the numbers, but not the exponentials. Convert answers to
scientific notation.
a. c. 1.0 x 10─14 =
3.25 x 10─5 1.0 x 10─14
8.8 x 10─4 b. 10─14 = 10─14 d. = = 4.3 x 10─4 2.4 x 103
e. [H+](6.7 x 10─12) = 1.0 x 10─14 ; [H+] =
f. (1.25 x 10─7) [OH─] = 1.0 x 10─14 ; [OH─] = ANSWERS
10─15 1a. (10─8)(10+2) = 10─6 1e. (10─3) [OH─] = 10─14 ; [OH─] = 10─11 2a. 10─17 3a. 3c. 6.5 x 10─8
3b. 5.0 x 10─13
3d. 8.79 x 10─3
324 x 10+12 = 3.24 x 10+14
5,260 = 5.26 x 10+3
4b. 0.010 = 1.0 x 10─2
4c. 0.0075 = 7.5 x 10─3 4a. 1b. 2b. 10─9 5b. 1/20 = 0.05 5e. 1/5 = 0.20 5f . 1/50 = 0.020 5c. 1/3 = 0.33 6a. (2.0 x 101) (3.0 x 10─11) = 6.0 x 10─10 7a. 2.5 x 10─13 10─2 1d. 10─5 2c. 10─3 5a. 1/2 = 0.50 6d. 5.0 x 10─7 1c. 5d. 1/4 = 0.25 5g. 1/8 = 0.125
6b. 5.0 x 10─19 5h. 1/80 = 0.0125
6c. 3.3 x 10─11 6e. (2.5 x 10─2)[OH─] = 10 x 10─15 ; [OH─] = 4.0 x 10─13
7b. 3.3 x 10─12 7c. 5.0 x 10─14 7d. (0.0125) [OH─] = 1.0 x 10─14 ; [OH─] = 1.0 x 10─14 = 0.80 x 10─ 12 = 8.0 x 10─13
1.25 x 10─2 © 2011 ChemReview.Net v. 2m Page 912 Module 29 — AcidBase Fundamentals 8a. 8c. 1.0 x 10─14 = 0.31 x 10─9
= 3.1 x 10─10
3.25 x 10─5
10─14 = 1.0 x 10─14 2.4 x 103 = 4.2 x 10─18 1.0 x 10─14 = 0.11 x 10─10
= 1.1 x 10─11
8.8 x 10─4 8b. 8e. 1.5 x 10─3 8d. 2.3 x 10─11 2.4 x 103 8f. (1.25 x 10─7)[OH─] = 1.0 x 10─14 ; [OH─] = 1.0 x 10─14 = 0.80 x 10─7 = 8.0 x 10─8
1.25 x 10─7
***** Lesson 29B: Kw Calculations: H+ and OH─
AcidBase Review
In Module 14, we considered acidbase neutralization calculations. In those problems, the
question was: if we have a known amount of an acid or base, can we find its stoichiometric
equivalent: the amount of the opposite that is needed to exactly neutralize the acid or base?
At the endpoint of a neutralization reaction, the particles of the acid and the base reactants
are exactly 100% used up, and the mixture is all products. These “reaction amount”
calculations for stoichiometric equivalents were solved using conversion stoichiometry.
Beginning in this module, we return to acids and bases to ask additional questions that are
important in both chemistry and biology. What is the nature of acid and base solutions
before they react? Which particles and ions are present? How are concentrated acid or
base solutions different from dilute solutions, and how do strong acids like hydrochloric
acid differ from weak acids such as vinegar that we frequently consume as food?
In most cases, we are interested in the behavior of acids and bases when they are dissolved
in an aqueous solution. Let’s start with the molecule that by definition is present in the
highest concentration in aqueous solutions: H2O . The Ionization of Water O A water molecule has 3 atoms  with oxygen in the
middle, two bonds, and a bent shape. H H At temperatures above absolute zero, the bonds in water bend and stretch. The molecules
also move at high average speeds, and they collide. A result of these motions is that at any
given time, at room temperature, for about one in 500 million molecules, one of the bonds
in liquid water is broken. The result is the formation of two ions, H+ and OH─, that also
move about in the water solution.
When a compound divides into ions that can move about freely, the reaction is termed
dissociation. When the initial compound has covalent character, the process can also be
termed ionization (forming ions). In practice, all compounds have some covalent
character, and the terms dissociation and ionization are often used interchangeably.
This separation of water into ions is reversible and can be represented by the equation
1 H2O(liquid) © 2011 ChemReview.Net v. 2m 1 H+(aq) + 1 OH─(aq) Page 913 Module 29 — AcidBase Fundamentals >99.999% unionized <0.001% ionized In pure (distilled) liquid water, the number of H+ and OH─ ions must be equal, since they
must be formed in a 1 to 1 ratio when water ionizes. The concentration of the H+ and OH─
ions must also be equal, since the equal moles of the two ions will be contained in the same
volume of solution. At room temperature, these ion concentrations are very small:
1.0 x 10─7 moles per liter for each. The concentration of the unionized water molecules is
very large in comparison: about 55 mol/L.
However, even at these low concentrations, small changes in the balance between H+ and
OH─ can have a large influence on reactions that are important in chemistry and biology. Water’s Ionization Constant: Kw
For the reversible ionization of water: H2O(liquid) H+(aq) + OH─(aq) the equilibrium constant expression could be written as:
K = [H+][OH─]
[H2O]
However, in aqueous solutions (those where water is the solvent), the concentration of the
nonionized water molecules is high (about 55 M) for the relatively dilute solutions used in
most lab experiments, and it does not change substantially during dilution or reactions.
In equilibrium constants, concentrations that remain close to constant during reactions are
generally included as part of the numeric value for K and omitted from the K expression.
The result is a simplified equilibrium constant expression (labeled Kw) for the relationship
between H+ and OH─ in an aqueous solution:
K (55 M) = Kw = [H+][OH─] At room temperature (25ºC), the value for Kw is 1.0 x 10─14. This small numeric value
(0.000 000 000 000 010) indicates that the reaction favors the reactants: very few water
molecules separate into ions.
As temperature increases, the molecules of water collide with higher energy, the bonds
bend and stretch more vigorously, and the bonds break more often. Of importance in
biology, at body temperature in mammals (37ºC), the [H+] in water is about 1.6 x 10─7, as
opposed to 1.0 x 10─7 at 25ºC.
However, for acidbase calculations in chemistry, you may assume that conditions are at
25ºC unless otherwise noted. In acidbase calculations, we will use this Kw equation:
Kw = [H+][OH─] = 1.0 x 10─14 at 25ºC This relationship between [H+] and [OH─ ] is an inverse proportion. If substances are
added to water to make one ion concentration increase, the other must decrease in the same
proportion: if one triples, the other must become 1/3 of its original concentration. © 2011 ChemReview.Net v. 2m Page 914 Module 29 — AcidBase Fundamentals AcidBase Terminology
In pure water, the concentration of H+ and OH─ ions must be equal. However, if acids or
bases are dissolved into water, this balance is upset.
By what are termed the classical definitions of acids and bases:
•
• An acid is a substance that ionizes in water and forms [H+] ions.
A base is a substance that ionizes in water and forms [OH─] ions. These are also called the Arrhenius definitions for acids and bases, after the Swedish
chemist Svante Arrhenius who first proposed the existence of electrically charged particles
(ions).
When acids or bases are added to water, the water continues to ionize slightly, and the Kw
equation will continue to predict the relationship between the [H+] and [OH─] ions. In an
acidic or basic solution, if either [H+] or [OH─] is known, the concentration of the other ion
can be calculated using the Kw equation.
A useful rule is: in calculations that include both [H+] and [OH─] in the WANTED and
DATA, write the Kw equation. We will call this the
Kw prompt: See [H+] and [OH─]? Write: Kw = [H+][OH─] = 1.0 x 10─14 Some problems will ask for the [H+] and [OH─] in acid or base solutions very
approximately, and those calculations can be done by mental arithmetic. Try this problem.
Q. In a solution with a [OH─] of about 10─2 M, what is the [H+]?
*****
Answer
WANT: [H+] DATA: [OH─] ≈ 10─2 M Prompt: See [H+] and [OH─]? Write: Kw = [H+][OH─] = 1.0 x 10─14 Substitute the known [OH─]:
[H+](~10─2) = 10─14 ; by inspection, [H+] must ≈ 10─12 M
In other problems, you will need to calculate the ion concentrations more precisely. Try
this calculation in your notebook, then check your answer below.
Q. If the [H+] in an aqueous solution is 5.0 x 10─3 M, find the [OH─].
*****
WANT:
DATA: [OH─] = ?
[H+] = 5.0 x 10─3 M
Kw = © 2011 ChemReview.Net v. 2m [H+][OH─] = 1.0 x 10─14 (Kw prompt) Page 915 Module 29 — AcidBase Fundamentals SOLVE: ? Solve the equation for the wanted symbol, then plug in the DATA.
= [OH─] = 1.0 x 10─14 = 1.0 x 10─14 = 0.20 x 10─11 = 2.0 x 10─12 M
[H+]
5.0 x 10─3 Recall that when solving K equations, units are omitted during calculations, but if the
WANTED unit is a concentration, the unit mol/L (M) is added to the answer.
*****
Kw Check
When using the Kw equation, do a check at the end: estimate [H+] times [OH─] (circled
above). The result must = 10.0 x 10─15 or 1.0 x 10─14 . Try multiplying the circled
values above in your head. Does the answer check? Practice: Do parts b and d of both, plus problem 3. Need more practice? Do more parts. 1. In these aqueous solutions, find the [H+] if the [OH─] is
a. 10─11 moles per liter b. 0.010 Molar c. 5.0 x 10─11 M d. 0.036 M 2. In these aqueous solutions, find the [OH─] if the [H+] is
a. 10─9 moles per liter b. 1.0 Molar c. 3.0 x 10─5 M d. 1.25 M 3. If 20.0 millimoles of OH─ ions are dissolved in 400. mL of solution, find [OH─] and
[H+]. ANSWERS
1 a. Since [H+][OH─] = 1.0 x 10─14 ; [H+](10─11) = 10─14 ; [H+] = 10─3 M b. 0.010 Molar = [OH─] ; = 1.0 x 10─2 M
[H+][OH─] = 1.0 x 10─14 ; [H+](1.0 x 10─2) = 1.0 x 10─14 ; [H+] = 1.0 x 10─12 M
c. 5.0 x 10─11 M = [OH─] ; WANT [H+]; the equation relating the two symbols is
[H+][OH─] = 1.0 x 10─14 . Solving for the wanted symbol:
[H+] = 1.0 x 10─14 = 1.0 x 10─14 = 0.20 x 10─3 = 2.0 x 10─4 M H+
[OH─]
5.0 x 10─11
d. 0.036 M = [OH─] ; WANT [H+]; the equation using those symbols is [H+][OH─] = 1.0 x 10─14 © 2011 ChemReview.Net v. 2m Page 916 Module 29 — AcidBase Fundamentals [H+] = 1.0 x 10─14 = 1.0 x 10─14 = 0.28 x 10─12 =
[OH─]
3.6 x 10─2
2 2.8 x 10─13 M H+ a. 10─9 moles per liter = [H+], want [OH─], the equation relating those symbols is
[H+][OH─] = 1.0 x 10─14 ; (10─9) [OH─] = 10─14 ; [OH─] = 10─5 M
b. 1.0 Molar = [H+], want [OH─] , that calls the Kw prompt:
write [H+][OH─] = 1.0 x 10─14 ; (1.0) [OH─] = 10─14 ; [OH─] = 1.0 x 10─14 M c. 3.0 x 10─5 M = [H+], want [OH─] , write [H+][OH─] = 1.0 x 10─14 ; [OH─] = 1.0 x 10─14 = 1.0 x 10─14 = 0.33 x 10─9 =
[H+]
3.0 x 10─5 3.3 x 10─10 M OH─ d. 1.25 M = [H+]; [OH─] = 1.0 x 10─14 = 1.0 x 10─14 = 0.80 x 10─14
[H+]
1.25
= 8.0 x 10─15 M OH─
3. WANTED = ? = [OH─] = mol OH─
L soln.
DATA: 20.0 mmol KOH = 400. mL soln. (two measures of same solution) (If you WANT a ratio, all of the DATA will be in equalities.)
SOLVE: ? mol OH─ = 20.0 mmol OH─ = 0.0500 mol OH─
L soln.
400. mL soln.
L soln (Since a prefix is an abbreviation for an exponential, the same prefix on the top and bottom can cancel.
Knowing [OH─], [H+] can be solved using the Kw prompt.)
[H+] = 1.0 x 10─14 = 1.0 x 10─14 = 0.20 x 10─12 = 2.0 x 10─13 M H+
[OH─] 5.00 x 10─2 ***** Lesson 29C: Strong Acid Solutions
Prerequisites: If you have difficulty with the calculations in this lesson, review Lesson 12B.
***** Definitions
By the classical (Arrhenius) definitions in chemistry, a compound that is a
• strong acid ionizes essentially 100% in water and forms an H+ ion;
strong base ionizes essentially 100% in water and forms an OH─ ion; • weak acid or base dissociates (ionizes) only partially when dissolved in water. • An H+ ion has one proton and no electrons, so the H+ ion is often referred to as a proton.
Acids can be classified as monoprotic (containing one hydrogen atom that can ionize) or
polyprotic (containing more than one). © 2011 ChemReview.Net v. 2m Page 917 Module 29 — AcidBase Fundamentals AcidBase States
Molecules that are acids and bases can react in a variety of ways.
For example, one type of acidbase reaction is the mixing of hydrogen chloride gas
and ammonia gas to form solid ammonium chloride.
HCl(g) + NH3(g) NH4Cl(s) However, most acidbase reactions are conducted in water, and unless a different state is
noted, in these lessons, if no state for an acidic or basic particle is shown, assume that the
state is aqueous (aq). Strong Acids
In chemistry, we often have a need for strong acid solutions. The strong acids most
frequently encountered are HCl, HNO3, and H2SO4.
• HCl and HNO3 are strong monoprotic acids. Both are highly soluble in water and
ionize essentially 100% to release one H+ ion.
Other strong monoprotic acids include HBr, HI, HClO4, and HMnO4, but because
these substances may undergo redox as well as acidbase reactions, they are used
less often for reactions that require a strong acid. • H2SO4 is a strong diprotic acid. When H2SO4 is dissolved in water, the first proton
ionizes essentially 100%, but the second ionizes only partially. The ionization of acids is more complex than the ionization of other ionic compounds in
part because the bond of the H in acids has a balance of ionic and covalent character.
Because of the mixed nature of bonds to hydrogen, to calculate ion concentrations in acidic
solutions, we need three sets of rules:
• one for strong acids (such as HCl and HNO3) that ionize completely, • one for weak acids, in which ionization goes to equilibrium, and • one for polyprotic acids (such as H2SO4) in which some H atoms ionize more easily
than others. Our rules for bases will parallel these acid rules. Let us start with the rules for solutions of
the strong acids HCl and HNO3. Ion Concentrations In Strong Acid Solutions
The concentration of an HCl or HNO3 solution is usually expressed in terms of an unionized acid formula, such as [HCl] = 0.25 M. However, the unionized formula does not
represent the particles that are actually present in the solution.
For example:
The gas hydrogen chloride (HCl) readily dissolves in water to form a solution of
hydrochloric acid. If 0.20 moles of HCl is dissolved per one liter of solution, the solution
concentration is written as “[HCl] = 0.20 M” based on how much HCl is added when
mixing the solution. © 2011 ChemReview.Net v. 2m Page 918 Module 29 — AcidBase Fundamentals However, there are no particles of HCl present in an “HCl solution.” Why? As HCl
dissolves in water, it immediately separates into ions:
1 HCl is used up 1 H+ is formed + 1 Cl─ is formed (goes ~ 100%) This reaction is reversible, but the right side is so strongly favored at equilibrium that in
water, essentially all of the HCl is converted to ions.
In a solution labeled [HCl] = 0.20 M, the actual concentrations of the particles are
[HCl] = 0 M ; [H+] = 0.20 M ; [Cl─] = 0.20 M ; and [H2O] ≈ 55 M
In addition, there is a very small amount of OH─ ion in solution (more on that later). Calculating Ion Concentrations In Strong Acid Solutions
The ionization of a strong acid parallels what happens to other ionic compounds that
separate into ions essentially 100% when dissolved in water (see Lesson 12B). The key rule
for calculations involving such compounds is:
For Substances Separating 100% Into Ions, Write the REC Steps
In calculations involving [ions], if a substance ionizes ~100%,
• R: Write the balanced ionization Reaction equation. After the equation, write • E: The Extent of the reaction (“goes ~100%”). Below each particle, write • C: The Concentration of the particle, based on the coefficient ratios. If a reaction goes to completion, the coefficients supply the mole ratios of reactants used up
and products formed. For the 100% dissociation reactions of strong acids, since all of the
reaction particles are in the same constant volume of solution, the coefficients also supply
the mole per liter reaction ratios.
Example: In 0.15 M HNO3 , what are the [ions]? Write the REC steps.
Rxn. and Extent:
Concentrations: 1 HNO3
^
0.15 M 0 M 1 H+ + 1 NO3─
^
^
0.15 M
0.15 M (goes ~ 100%) Try the following problem, and then check your answer below.
Q. In a 0.45 M HCl solution, write the
a. [HCl]as mixed
b. [H+]in soln.
c. [Cl─]in soln.
*****
Answer: Since HCl is a strong acid, in water it ionizes ~100%. To find ion concentrations
for substances that ionize ~100%, write the REC steps.
Rxn. and Extent: 1 HCl used up
^
Conc.:
0.45 M 0 M 1 H+ formed + 1 Cl─ formed (goes ~ 100%)
^
^
0.45 M
0.45 M The bottom row shows the WANTED concentrations. The [HCl] as mixed was 0.45 M,
but in solution is 0 M. © 2011 ChemReview.Net v. 2m Page 919 Module 29 — AcidBase Fundamentals In some problems, to find [ions], the [HCl] or [HNO3] as mixed will need to be solved first.
Try
Q. If 0.030 moles of HNO3 is mixed with water to form 150 mL of solution, find
b. [H+]in soln.
c. [NO3─]in soln.
a. [HNO3]as mixed
*****
Answer (In problems with parts, you may list either the WANTED or the DATA first.)
a. WANT: ?= mol HNO3
L soln DATA: 0.030 mol HNO3 = 150 mL soln. (two measures; same soln.) (When a ratio unit is WANTED, all DATA will be in equalities.)
SOLVE: (To review molarity calculations, see Lesson 11C.)
? = mol HNO3 = 0.030 mol HNO3 ● 1 mL = 0.20 M HNO3
L soln.
150 mL soln.
10─3 L b,c. WANTED: [H+]in soln. and [NO3─]in soln.
(HNO3 is a strong acid: it ionizes ~100%. To find [ions], write the REC steps.)
Rxn. and Extent:
Concentrations: 1 H+ + 1 NO3─
^
^
0.20 M
0.20 M 1 HNO3
^
0.20 M 0 M (goes ~ 100%) Practice A
1. In a 0.50 M solution of nitric acid (HNO3), what will be the
a. [H+] ?
b. [NO ─] ?
3 c. [H3O+] ? 2. 7.30 grams of HCl is dissolved in water to make 250 mL of solution. Find the
a. moles of HCl dissolved in the solution, per liter.
b. [H+] = c. [Cl─] = d. [H3O+] = Why [Acid] Determines [H+]
A strong acid solution is a mixture of the acid and water. It will contain four particles.
• The strong acid is present as two ions: an H+ ion and the acid’s anion.
• Water is the largest component in aqueous solutions, present primarily as nonionized H2O. Water also contains the small amounts of H+ and OH─ that form
due to its ionization.
When a strong acid is dissolved in water, H+ is contributed by both the acid and the water.
However, if the strong acid is mixed in any significant concentration, the share of H+ ions
contributed by the water can be ignored. © 2011 ChemReview.Net v. 2m Page 920 Module 29 — AcidBase Fundamentals Why? Consider a 0.20 M HCl solution. The strong acid contributes 0.20 moles of H+ ions
per liter to the water. Before the acid was added, the water contained only 10─7 moles of
H+ ions per liter. If we examine the uncertainty in these two amounts:
H+ from the acid, with doubt in the hundredth’s place, compares to
0.0000001 M H+ initially in the water, with doubt in a much higher place.
0.20 M This is one indication that any initial H+ contribution from the water’s ions is too small to
be significant. The acid ionization is the dominant reaction. For this reason, the rule is:
In an acid solution, use acid ionization rules to find [H+] The HCl and HNO3 Quick Rule
REC steps show concentrations for all ions formed when substances ionize 100%.
However, in many strong acid calculations, only [H+] is wanted, and you can use this
Quick rule: [HCl]mixed or [HNO3]mixed = [H+]in solution
Example: In a solution labeled 0.35 M HCl , [H+]in soln. = 0.35 M
When strong monoprotic acids dissolve in water, the ratio of the [acid as mixed but used
up] and the [H+] that forms in solution is always 1 to 1. H+ and H3O+ Ions  Equivalent
The proton released by an acid is often represented as H+, but in aqueous solutions the
proton nearly always is found attached to a water molecule, forming a hydronium ion
(H3O+). This reaction can be represented as
1 H O+
(goes ~ 100%)
1 H+ + 1 H O
2 3 Textbooks often show acids in water forming H+ in some reactions and H3O+ in others. In
calculations involving acids in aqueous solutions, the symbols H+ and H3O+ in most cases
are considered to be equivalent. When H3O+ is encountered in calculations, use the
H3O+ Prompt. If you see H3O+, write: H3O+ = H+ Summary: Acid Rules
1. See [H+] and [OH─] ? Write: Kw = [H+][OH─] = 1.0 x 10─14
2. Strong monoprotic acids ionize ~100% to form H+. For [ions], write the REC steps.
3. Quick rule: [HCl or HNO3]mixed = [H+]in soln. 4. See H3O+ ? Write: H3O+ = H+ 5. In an acid solution, use acid ionization rules to find [H+]. © 2011 ChemReview.Net v. 2m Page 921 Module 29 — AcidBase Fundamentals Practice B: Memorize the rules above, then do these problems. 2. In 2.0 M HCl, a. [H+] = b. [H3O+] = a. [H3O+] = 1. In 0.25 M HNO3 b. [H+] = ANSWERS
Practice A
1a,b. For problems involving HCl or HNO3 and [ions], write the REC steps for 100% ionization.
Rxn. and Extent:
Conc.: 1 H+ + 1 NO3─
^
^
0.50 M
0.50 M 1 HNO3
^
0.50 M 0 M 1c. [H3O+] = [H+] = 0.50 M 2a. WANT: ? mol HCl = [HCl]as mixed
L soln (goes ~ 100%) (it helps to label units with the symbols
used in rules and equations) DATA: 7.30 g HCl = 250 mL soln (equivalent: two measures; same soln.) 36.5 g HCl = 1 mol HCl (grams prompt) You want moles over liters. The data includes grams, moles and mL. Use conversions. WANT a
ratio? Start with a ratio.
SOLVE: ? mol HCl = 7.30 g HCl ● 1 mol HCl ●
L soln 250 mL soln 36.5 g HCl 1 mL = 0.80 M HCl = [HCl]mixed
10─3 L When solving in parts, label answers with symbols from rules that may be needed in later parts.
2b,c. For problems involving HCl or HNO3 and [ions], solve by the REC steps.
Rxn. and Extent:
1 HCl
1 H+ + 1 Cl─
( goes ~ 100%)
^
^
^
0.80 M
0.80 M
Conc.:
0.80 M 0 M
[HCl]mixed = 0.80 M from part a = [H+]in soln = [Cl─]in soln d. [H3O+] = [H+] = 0.80 M Practice B
Quick rule: In HCl or HNO3 , for [H+], use [HCl or HNO3]mixed = [H+]in soln.
1. In 0.25 M HNO3
2. In 2.0 M HCl, a. [H+] = 0.25 M b. [H3O+] = [H+] = 0. 25 M a. [H3O+] = [H+] = 2.0 M b. [H+] = 2.0 M ***** © 2011 ChemReview.Net v. 2m Page 922 Module 29 — AcidBase Fundamentals Lesson 29D: The [OH─] in Strong Acid Solutions
The Impact of a Strong Acid on the [OH─ ] in Water
The ionization of an acid in water affects the other reaction occurring in an acid solution:
the reversible “autoionization” of water.
1 H+(aq) + 1 OH─(aq) 1 H2O(liquid) (goes slightly) Adding acid to water shifts this equilibrium in accord with Le Chatelier’s Principle.
Adding acid increases the [H+]. The equilibrium shifts to the opposite side from the H+,
decreasing the [OH─] . The value of the [OH─] after the shift can be calculated by the
equilibrium constant for water’s ionization:
Kw = [H+][OH─] = 1.0 x 10─14 at 25ºC When solving for [H+] and [OH─] in an acid solution, let us add to Rule
5. In an acid solution, first use acid ionization rules to find [H+], then Kw to find [OH─].
Use the rule for this problem.
Q. In a 0.40 M HCl solution, find a. [H+] b. [Cl─] c. [OH─] * * ** *
a,b. To find [ions] in a strong acid solution, write the REC steps.
Rxn. and Extent: 1 HCl
^
Conc.:
0.40 M 0 M
c. 1 H+ +
^
0.40 M 1 Cl─
(goes ~ 100%)
^
0.40 M WANT [OH─].
In an acid solution, first use acid rules to find [H+], then Kw to find [OH─].
Part a found [H+] = 0.40 M
Kw = To find [OH─], use Kw . [H+][OH─] = 1.0 x 10─14 ; ? = [OH─] = 1.0 x 10─14 = 1.0 x 10─14 =
[H+]
0.40 2.5 x 10─14 M Kw check: [H+] x [OH─] (circled above) must ≈ 10.0 x 10─15 or 1.0 x 10─14. Check. © 2011 ChemReview.Net v. 2m Page 923 Module 29 — AcidBase Fundamentals Summary: AcidBase Rules
1. See [H+] and [OH─] ? Write: Kw = [H+][OH─] = 1.0 x 10─14
2. Strong monoprotic acids ionize ~100% to form H+. For [ions], write the REC steps.
3. Quick rule: [HCl or HNO3]mixed = [H+]in soln. 4. See [H3O+] ? Write: [H3O+] = [H+]
5. In acid solutions, use acid ionization rules to find [H+], then Kw to find [OH─]. Practice
1. From memory, write the 5 numbered acidbase rules learned so far in this module.
2. In a solution labeled 10─3 M HNO3, what will be the
a. [H+] ? b. [NO3─] ? c. [H3O+] ? d. [OH─] ? 3. In a solution labeled 0.020 M HCl,
a. Write the balanced equation for the ionization of this strong acid.
b. Which side will be favored in this ionization: products, or reactants?
c. What is the mole to mole ratio between HCl used up and H+ formed?
d. What is the mol/L ratio between HCl used up and H+ formed?
e. In the solution, what will be the
i. [H+] ? ii. [Cl─] ? iii. [OH─] ? iv. [H3O+] ? ANSWERS
2 a,b. For problems involving HCl or HNO3 and [ions], solve by the REC steps.
Rxn. and Extent:
Conc.: 1 HNO3
^
10─3 M 0 M 1 H+ + 1 NO3─ ( goes ~ 100%)
^
^
10─3 M
10─3 M 10─3 M = [H+] = [H3O+] 2c. To find [OH─ ] in an acid solution, use acid rules to find [H+], then Kw to find [OH─ ]. 2d. Since [H+] = 10─3 M , and [H+][OH─] = 1.0 x 10─14 ; [OH─] = 10─11 M.
3a,b. Equation: 1 HCl(g) 1 H+(aq) + 1 Cl─(aq) (goes 100%) Products favored c. 1 mol HCl used up = 1 mole H+ formed.
d. Since all moles are found in the same liters of solution, the mole and mol/L ratios are the same: 1 to 1. © 2011 ChemReview.Net v. 2m Page 924 Module 29 — AcidBase Fundamentals 1 mol/L HCl used up = 1 mol/L H+ formed, which can be written: [HCl]used up = [H+]formed
e. [HCl]as mixed = 0.020 M = [H+]in soln. = [Cl─]in soln.
[OH─] ? In an acid solution, use acid rules to find [H+ ] (done above), then use Kw to find [OH─ ]. Since [H+] = 0.020 M = 2.0 x 10─2 M , and [H+][OH─] = 1.0 x 10─14 ;
[OH─] = 1.0 x 10─14 = 1.0 x 10─14
[H+]
2.0 x 10─2 = 0.50 x 10─12 = 5.0 x 10─13 M Kw check: [H+] x [OH─] (circled above) must estimate to 10.0 x 10─15 or 1.0 x 10─14
[H+] x [OH─] = 2 x 5 = 10; 10─2 x 10─13 = 10─15 , combined = 10 x 10─15. Check.
iv. [H3O+] = [H+] = 0.020 M = 2.0 x 10─2 M
***** Lesson 29E: Strong Base Solutions
Strong Hydroxide Bases
Strong bases, by the classical (Arrhenius) definitions, ionize essentially 100% in water to
form OH─ ions. The substances most frequently used to make strong base solutions are the
water soluble ionic solids sodium hydroxide (NaOH) and potassium hydroxide (KOH). In
water,
1 NaOH(s) 1 Na+(aq) + 1 OH─(aq) 1 KOH(s) 1 K+(aq) + 1 OH─(aq) (goes ~ 100%)
(goes ~ 100%) There are other types of strong bases, but their reactions can be more complex. For now,
we will limit our attention to the strong bases that are alkali metal hydroxides. [Ions] In Strong Base Solutions
As with strong acids, if a solution of a strong base is labeled [NaOH] = 0.15 M, this
represents how the solution is mixed, but not the particles present in the solution. Because
NaOH ionizes ~100% in water, what is actually present in “0.15 M NaOH” is
zero M NaOH, 0.15 M Na+, and 0.15 M OH─.
The rules for strong bases are similar to those for strong acids.
● NaOH and KOH ionize ~100% to form OH─. For [ions], write the REC steps.
● If only [OH─] is needed, use the quick rule: [NaOH or KOH]mxd = [OH─]in soln.
● In a base solution, use base rules to find [OH─], then Kw to find [H+]. © 2011 ChemReview.Net v. 2m Page 925 Module 29 — AcidBase Fundamentals We will add these to our rules that, when committed to longterm memory, simplify acidbase problem solving. AcidBase Fundamentals
1. See [H+] and [OH─] ? Write: Kw = [H+][OH─] = 1.0 x 10─14
2. Strong monoprotic acids ionize ~100% to form H+. Alkali metal hydroxides
ionize 100% to form OH─. For [ions], write the REC steps.
3. Quick rules: [HCl or HNO3]mixed = [H+]in soln.
[NaOH or KOH]mixed = [OH─]in soln. 4. See [H3O+] ? Write: [H3O+] = [H+]
5. In acid solutions, use acid ionization rules to find [H+], then Kw to find [OH─].
6. In base solutions, use base rules to find [OH─], then Kw to find [H+].
Using the new rules above, try this problem.
Q. In a solution labeled 0.0030 M NaOH,
a. What three ions are present in the solution?
b. What is the concentration of each ion?
*****
Answer
a. NaOH ionizes to form Na+ and OH─ ions. Water ionizes to form H+ and OH─ .
The three ions are Na+, OH─, and H+.
b. For NaOH or KOH and [ions], write the REC steps for ~100% ionization.
Rxn. and Extent: 1 NaOH(aq)
Conc.: ^
0.0030 M 0 M 1 Na+(aq)
^
0.0030 M + 1 OH─(aq) (goes ~ 100%) ^
0.0030 M which means [NaOH]mixed = 0.0030 M = [OH─]in soln. = [Na+]in soln.
To find the [H+] in a base solution, first use the base rules to find [OH─] (done),
then use Kw to find [H+].
[H+] = 1.0 x 10─14 = 1.0 x 10─14 = 0.33 x 10─ 11 =
[OH─] 3.3 x 10─12 M H+ 3.0 x 10─3 (Kw check: [H+] x [OH─] = 3 x 3.3 ≈ 10; 10─3 x 10─12 = 10─15, combined ≈ 10 x 10─15) © 2011 ChemReview.Net v. 2m Page 926 Module 29 — AcidBase Fundamentals Practice: Learn the AcidBase Fundamental rules above, then try these. 1. In a solution labeled 10─1 M NaOH,
a. write the balanced equation for the ionization of this strong base.
b. Which side is favored in this ionization: products, or reactants?
c. Write the i. [OH─] ii. [Na+] iv. [H3O+] iii. [H+] 2. In a solution labeled 5.0 x 10─3 M KOH, find the
a. [OH─] b. [K+] d. [H3O+] c. [H+] 3. If 5.00 millimoles NaOH are dissolved in 0.250 L soln, find
a. [OH─] b. [H+] c. [H3O+] ANSWERS
1 Na+ + 1 OH─ 1a. 1 NaOH 1b. Strong bases ionize 100%, this ionization favors the products.
1c. i, ii. 1 NaOH
^
─1 0 M
10 1 Na+ + 1 OH─
^
^
─1 M
─1 M
10
10 ( goes ~ 100%) [NaOH]as mixed = 10─1 M = [OH─]in soln. = [Na+] in soln.
iii,iv. [H+] ? To find [H+] in a base solution, first find [OH─] (above), then [H+] using
Kw. = [H+][OH─] = 1.0 x 10─14
[H+] = 1.0 x 10─14 = 1.0 x 10─14 = 10─13 M = [H+] = [H3O+]
[OH─]
10─1
2a,b. For NaOH or KOH, to find [ions], write the REC steps for ~100% ionization.
R and E:
Conc.: 1 KOH
^
5.0 x 10─3 0 M 1 K+
+
^
5.0 x 10─3 M 1 OH─
( goes ~ 100%)
^
5.0 x 10─3 M [KOH]mixed = 5.0 x 10─3 M = [OH─]in soln. = [K+] in soln.
c. [H+] = ? [OH─] is known. The relationship between those two is: [H+][OH─] = 1.0 x 10─14
[H+] = 1.0 x 10─14 = 1.0 x 10─14 = 0.20 x 10─11 = 2.0 x 10─12 M H+
[OH─]
5.0 x 10─3 © 2011 ChemReview.Net v. 2m Page 927 Module 29 — AcidBase Fundamentals d. [H3O+] ? = [H+] = 2.0 x 10─12 M Note that each of the basic solutions above contain both OH─ and H+, but in solutions of bases, [OH─] is
always higher than [H+].
3a. WANTED = ? = [OH─] = mol OH─
L soln. First find [NaOH], then use [NaOH] = [OH─] DATA: 5.00 mmol NaOH = 0.250 L soln. (two measures of same solution) SOLVE: ? mol NaOH = 5.00 mmol NaOH · 10─3 mol = 2.00 x 10─2 mol = [NaOH] = [OH─]
L soln.
0.250 L soln.
L
1 mmol [H+] = 1.0 x 10─14 = 1.0 x 10─14 = 0.50 x 10─12 = 5.0 x 10─13 M H+ = [H+] = [H3O+]
[OH─]
2.00 x 10─2
***** Lesson 29F: The pH System
Prerequisites: If you cannot do all of these problems correctly, do Lesson 27D before
starting this lesson. Use a calculator. Answers are at the end of this lesson.
2. log (2.0 x 10─12) = 1. Log 10─4 = Answer the following in scientific notation.
3. If log x = ─ 9.2 , x = 4. 10─12.3 = ***** Fundamentals of pH
pH is a measure of the proton concentration in an aqueous solution. The pH is a number
between ─2 and 16 that indicates the acidity or basicity (baySISahtee) of the solution.
0
7
14
Acidic
Neutral
Basic
The precise definition of pH is pH ≡ ─ log { [H+] / (mol/L) }
Dividing [H+] by the units of concentration is needed to comply with the mathematical rule
that logarithms cannot be taken of quantities with units. However, to speed calculations,
we often use a simplified definition in which
pH is defined as the negative log of the hydrogen ion concentration: pH ≡ ─ log [H+]
If we rearrange the simplified definition to solve for [H+], the result is: [H+] ≡ 10─pH
It may help to remember pH as the “power of H;” the number that is after the minus sign
when [H+] written as a negative power of 10. © 2011 ChemReview.Net v. 2m Page 928 Module 29 — AcidBase Fundamentals In calculations that involve pH, we will use these definitions as the
pH Prompt. If you see pH, write: pH ≡ ─ log [H+] and [H+] ≡ 10─pH
The pH prompt should be committed to memory.
When using these simplified definitions,
If a pH value is calculated, it is not given a unit, but when a concentration is calculated
from a pH, the unit mol/L (M) must be added to the answer.
It is also important to know that:
• Pure water has a pH of 7.0 at 25ºC. • In a pHneutral solution at 25ºC, [H+] = [OH─] = 1.0 x 10─7 M; and pH = 7. • In acidic solutions, the pH is less than 7. In basic solutions, the pH is greater than 7. • Lower pH means a higher acidity. Higher pH means a higher basicity. • The further from 7 is the pH, the stronger is the acidity or basicity of the solution. • On a base 10 logarithmic scale, changing a number by a factor of 10 changes the log
by one. • Since pH is a negative log, increasing [H+] by a factor of 10 lowers the pH by one. Why Use the pH Scale?
The pH scale allows us to report the acidity of aqueous solutions without using exponents.
For nonchemists, saying that in my aquarium “the pH is 7.5” involves easier numbers than
“the [H+] is 3.2 x 10─8 mol/L,” though both of those statements have the same meaning.
For chemists, pH is a quick way to convey solution acidity. A. Whole Number pH
In aqueous solutions, if either the [H+] or the [OH─] or the pH is known, the other two
values can be calculated.
When pH values are whole numbers, this can be done by inspection. Try these examples:
Q1. In an aqueous solution, if [H+] = 10─1 M, what is the pH?
* ** * *
[H+] ≡ 10─pH = 10─1 M, so pH = 1. The solution is acidic.
Q2. If [H+] = 0.001 M, what is the pH?
* ** * *
[H+] = 0.0001 M = 10─3 M = 10─pH, so pH = 3, and the solution is acidic.
Q3. If the pH = 9, what is the [H+]? Is the solution acidic or basic?
* ** * * © 2011 ChemReview.Net v. 2m Page 929 Module 29 — AcidBase Fundamentals The [H+] = 10─9 M, and the solution is basic. Why? Since [H+] = 10─9 M,
[OH─] must equal 10─5 M (based on Kw). The [OH─] is larger than the
[H+], and that is the definition of a basic solution. Any pH > 7 is basic. Practice A: Whole Number pH
Do these calculations in your head. Do them all. Check answers at the end of the lesson.
1. Write the pH prompt from memory.
For aqueous solutions, if
2. [H+] = 10─4 M, a. [OH─] = 3. pH = 8 , a. [H+] = b. pH =
b. [OH─] = 4. [OH─] = 10─3 M, a. [H+] = 6. [H+] = 10 M, a. [OH─] = c. Is the solution acidic or basic? b. pH = 5. [OH─] = 1.0 x 10─14 M, a. [H+] = c. Is the solution acidic or basic? c. Is the solution acidic or basic? b. pH =
b. pH = c. Acidic or basic solution?
c. Acidic or basic solution? B. From [H+] to Decimal pH
If the pH or the exponent of the hydrogen ion concentration is not a whole number, you can
convert between [H+], [OH─], and pH using a calculator. Try this example.
Q. For an aqueous solution, if [H+] = 5.0 x 10─4 M, what is the pH?
*****
WANT: pH, know [H+].
pH ≡ ─ log [H+] Write the equation that solves for pH from [H+]: = ─ log (5.0 x 10─4) = ? Solve using a calculator, round your answer to tenths, and then check it below.
*****
This is a brief review of log calculations. For detail, see Lesson 27D.
• On a standard TI–type calculator, try: 5.0 EE 4 +/ • An RPN calculator might use: 5.0 • On one type of graphing calculator: E log +/ 4 enter log
() log 5.0 EE +/+/ () 4 ) enter For your calculator, either read your calculator manual (the best approach) or experiment
until you get the following answer. Write down the key sequence that works.
pH = ─ log (5.0 x 10─4) = 3.30
To simplify calculator use, you may want to find the pH without the minus sign in front of
the log (but you must include the minus sign of the exponential term) , then change the © 2011 ChemReview.Net v. 2m Page 930 Module 29 — AcidBase Fundamentals sign manually. For any solution with a [H+] less than 1.0 M, the pH will be positive. If the
[H+] is greater than 1.0 M, which can occur only in relatively concentrated strong acid
solutions, the pH will be a negative number. Checking pH Calculations
Does the above answer make sense?
For [H+]: 1.0 x 10─4 < 5.0 x 10─4, < 10.0 x 10─4 ≡ 1.0 x 10─3,
pH = 4 pH = ? pH = 3 Based on the above, the pH should be between 4 and 3, which 3.30 is.
Here’s another way to estimate and check pH: Round the pH up to the next highest whole
number. That should be the number after the minus sign in the exponential term of the
[H+] when [H+] is written in scientific notation (as it should be).
For this problem, pH = 3.30 rounds up to 4. The [H+] exponential term in scientific
notation should therefore be 10─4, and it is: 5.0 x 10─4.
When complex operations are done on a calculator, estimate the answer – or do the
calculation two ways on the calculator – as a check on your calculator operation. Rounding pH Calculations
Mathematically, the statistical basis for using significant figures to convey uncertainty does
not directly apply to logarithmic functions. In these lessons, we will use these rules for pH
calculations.
• When a [H+] is written as simply a power of 10 (with no significand in front), the pH
is written as a whole number;
Example: [H+] = 10─4 M, pH = 4 ; • When [H+] is written in scientific notation, round the pH so that the number of digits
in the significand equals the number of digits after the decimal in the pH.
Examples: [H+] = 5 x 10─4 M, pH = 3.3 ; [H+] = 5.1 x 10─12 M, pH = 11.29 • Due to the inherent uncertainty of the Kw values on which pH is based, calculated pH
values will be rounded to have at most two digits past the decimal. Practice B: From [H+] to Decimal pH
Do the odd problems. Save the evens for your next practice session. Check answers at the
end of the lesson as you go.
1. Using the quickcheck method above, estimate the pH for these questions below:
a. Problem 2: pH ≈ © 2011 ChemReview.Net v. 2m b. Problem 3: pH ≈ c. Problem 5: pH ≈ Page 931 Module 29 — AcidBase Fundamentals Use your calculator on the following.
2. Find pH when [H+] = 2.1 x 10─3 M.
3. If [H3O+] = 8.2 x 10─11 M, what is the pH?
4. If [OH─] = 2.0 x 10─4 M, what is the pH?
5. In a 0.040 M HCl solution, solve for the pH.
6. In a 0.0030 M KOH solution, what is the pH? C. From Decimal pH to [H+]
Let’s try a problem where a nonwholenumber pH is known and the [H+] is WANTED.
Example: If the pH of a solution is 9.70, what is the [H+] ?
The equation that finds [H+] from pH is: [H+] ≡ 10─pH [H+] therefore equals 10─9.70 M. That’s mathematically correct, but in chemistry it is
preferred to express concentrations in scientific notation, so that the exponent of 10 is a
whole number. Convert 10─9.70 to scientific notation using your calculator, and then
check your answer below.
*****
• On a standard TI–type calculator, try: 9.70 +/ 10x
+/ 10x • A reverse Polish (RPN) calculator might use: 9.70 • On a graphing calculator, try: 10x () 9.70 ) enter or 10 ^ () 9.70 enter For additional help, see Lesson 27D.
that replicates this result: Whatever method is used, write down the sequence [H+] = 10─9.70 M = 2.0 x 10─10 M
Does this answer make sense? [H+] = 10─9.70 M is close to [H+] = 10─10 M. Since
pH = 9.70 is a bit more acidic than pH = 10, the [H+] should be a bit higher than 10─10 M,
and it is: 2.0 x 10─10 M.
Another way to check your answer is to use this pH estimating step: Round the pH up to
the next highest whole number. This should match the number after the minus sign of the
exponent for [H+] in scientific notation.
For this problem, pH = 9.70 rounds up to 10, so the [H+] exponent should be 10─10.
The answer is 2.0 x 10─10. Check! © 2011 ChemReview.Net v. 2m Page 932 Module 29 — AcidBase Fundamentals Practice C: From Decimal pH to [H+]
Use a calculator. Save a few problems for your next practice session.
1. If pH = 5.5 a. Estimate the [H+] 2. If pH = 8.20, b. [H+] = a. Estimate the [H+]
b. [H+] =
c. [OH─] = Work Problems 37 in your problem notebook.
3. In an HCl solution of pH = 3.60, find
a. [H+] b. [OH─] 4. If pH = 1.7, [H+] = c. [Cl─] d. [HCl] 5. If pH = 7.22, [H+] = 6. If pH = 12.5, [H+] = 7. If pH = ─0.50, [H+] = 8. Which solution in problems 37 is the most acidic? D. Calculations With pOH
(If you are asked to solve problems that include pOH, do this section.)
The pOH is the negative log of the hydroxide ion concentration. The pOH can be solved
using the same math as pH.
In problems that include pOH, our rule will be the
pOH prompt: See pOH?
Write pOH ≡ ─ log[OH─] and [OH─] ≡ 10─pOH and pH + pOH = 14.00 The pOH is often calculated by first finding the pH and then applying the pH + pOH
relationship.
The 14.00 represents the value of at Kw at 25ºC. At other temperatures, the relationship is
pH + pOH = pKw = ─log(Kw) pH Summary: Add these to the rules for acidbase fundamentals.
7. See pH ? Write: pH ≡ ─ log [H+] and [H+] = 10─pH 8. See pOH ? Write: pOH ≡ ─ log[OH─] and [OH─] = 10─pOH
and pH + pOH = 14.00 (at 25ºC) . © 2011 ChemReview.Net v. 2m Page 933 Module 29 — AcidBase Fundamentals Practice D: pOH Calculations
Memorize the three pOH rules above, and then solve problems 14 by inspection.
1. In an aqueous solution, if [OH─] = 10─5 M,
a. [H+] = b. pH = c. pOH = d. Is the solution acidic or basic? 2. In an aqueous solution, if [H+] = 10─4 M,
a. [OH─] = b. pH = c. pOH = d. Is the solution acidic or basic? 3. In an aqueous solution, if the pH = 12 ,
a. [H+] = b. [OH─] = c. pOH = d. Is the solution acidic or basic? 4. In an aqueous solution, if the pOH = 8
a. pH = b. [H+] = c. [OH─] = d. Is the solution acidic or basic? Use a calculator for Problems 56. Work these in your notebook.
5. 2.00 grams of NaOH is dissolved in water to make 750. mL of solution.
b. [OH─]in solution = a. [NaOH]as mixed =
c. [Na+]in soln. = d. [H+] = e. pOH = f. pH = 6. The pOH in a KOH solution is 1.5 . What is the [KOH]? ANSWERS
Pretest: 1. ─ 4 2. ─ 11.7 3. 6.3 x 10─10 4. 5.0 x 10─13 Practice A: Whole Number pH
a. [OH─] = 10─10 M 2. If [H+] = 10─4 M,
3. If pH = 8 , b. [OH─] = 10─6 M a. [H+] = 10─8 M 4. If [OH─] = 10─3 M, b. pH = 4 a. [H+] = 10─11 M b. pH = 11 c. Acidic
c. Basic
c. Basic 5. If [OH─] = 1.0 x 10─14 M,
a. [H+] = 1.0 M b. pH = ? [H+] = 1.0 = 10─pH = 100 = 10─0 ; pH = 0 c. Acidic. pH = 0 which is less than 7; [H+] is higher than [OH─].
6. If [H+] = 10 M, a. [OH─] = 10─15 M (pH will be lower than zero when [H+] > 1.0 M) © 2011 ChemReview.Net v. 2m b. [H+] = 101 = 10─pH = 10─(─1) ; so pH = ─1
c. Highly acidic Page 934 Module 29 — AcidBase Fundamentals Practice B: From [H+] to Decimal pH
1. a. Problem 2: pH = 2.?  rounds up to 3 Using the quickcheck method, b. Prob. 3: pH = 10.? (rounds up to 11) c. Prob. 5: 0.040 M HCl = 4 x 10─2 M H+ ; so pH = 1.?
2. pH prompt: see pH, write pH ≡ ─ log [H+] and [H+] ≡ 10─pH Since pH is WANTED, use ? = pH ≡ ─ log [H+]
pH = ─ log (2.1 x 10─3) = 2.68
3. (Estimate was 2.? in answer 1a. Check.) If [H3O+] = 8.2 x 10─11, what is the pH?
See pH, write pH ≡ ─ log [H+] and [H+] ≡ 10─pH pH is WANTED, use ? = pH ≡ ─ log [H+]
4. See pH, write pH ≡ ─ log [H+] and [H3O+] = [H+] = ─ log (8.2 x 10─11) = 10.09 (checks above) [H+] ≡ 10─pH WANT pH, need [H+] to find pH, but are given [OH─].
See H+ and OH─ ? Write: Kw = [H+][OH─] = 1.0 x 10─14 Use Kw to find [H+]: [H+] = 1.0 x 10─14 = 1.0 x 10─14 = 0.50 x 10─10 = 5.0 x 10─11 M H+
[OH─]
2.0 x 10─4
(do the Kw check.)
Then ? = pH ≡ ─ log [H+]
5. See pH, write pH ≡ ─ log [H+] and WANT pH. Need [H+] first.
For pH:
6. pH = ─ log (5.0 x 10─11) = 10.30 (pH rounds up to 11; check.) [H+] ≡ 10─pH For quick [H+]in soln.: [HCl]mixed = 0.040 M = [H+]in soln. pH ≡ ─ log [H+] = ─ log (0.040) = ─ log (4.0 x 10─2) = 1.40 (rounded up = 2. check.) pH ≡ ─ log [H+] and [H+] ≡ 10─pH Need [H+] to find pH, but given is [KOH]. Quick: [KOH]mixed = 0.0030 M = [OH─] To find [H+],
[H+] = 1.0 x 10─14 = 1.0 x 10─14 = 0.33 x 10─11 = 3.3 x 10─12 M H+
[OH─]
3.0 x 10─3
pH = ─ log (3.3 x 10─12) = 11.48 © 2011 ChemReview.Net v. 2m (Base solution, basic pH; 11.48 rounded up = 12. Check.) Page 935 Module 29 — AcidBase Fundamentals Practice C: From Decimal pH to [H+]
1. See pH, write If pH = 5.5
a. Estimate [H+] pH ≡ ─ log [H+] and [H+] ≡ 10─pH pH = 5.5 rounded up is 6; [H+] should be ? x 10─6 M b. The equation that finds [H+] from pH is: [H+] = 10─pH ; [H+] = 10─5.5 =
pH ≡ ─ log [H+] and ←compare ↑ 3 x 10─6 M [H+] ≡ 10─pH 2. See pH? Write: 2a. Estimate: pH = 8.20 rounds up to 9, [H+] should = ?.? x 10─9 M
WANT [H+] from pH, use [H+] = 10─pH ; [H+] = 10─8.20 = 6.3 x 10─9 M H+
WANT [OH─] . Know [H+] and pH. Checks vs. pH . Kw relates [H+] and [OH─ ]: [H+][OH─] = 1.0 x 10─14 ; ? = [OH─] = 1.0 x 10─14 = 1.0 x 10─14 = 0.16 x 10─5 = 1.6 x 10─6 M OH─
[H+]
6.3 x 10─9
3. pH ≡ ─ log [H+] and [H+] ≡ 10─pH See pH? Write [H+] ≡ 10─pH a. [H+] = ? = 10─3.60 = 2.5 x 10─4 M b. [OH─] = ? Knowing [H+], to find [OH─ ], use Kw .
? = [OH─] = 1.0 x 10─14 = 1.0 x 10─14 = 0.40 x 10─10 = 4.0 x 10─11 M OH─
[H+]
2.5 x 10─4
c, d. [HCl]mixed = [H+]soln = [Cl─]soln
4. If pH = 1.7, 6. pH = 12.5, [H+] ≡ 10─pH
7. pH = ─0.50, [H+] ≡ 10─pH 2.5 x 10─4 M = [Cl─] = [HCl] = 10─1.7 = 2 x 10─2 M [H+] ≡ 10─pH 5. If pH = 7.22, [H+] ≡ 10─pH From Part a, [H+] = = 10─7.22 = 6.0 x 10─8 M
= 10─12.5 = 3 x 10─13 M
= 10─(─0.50) = 100.50 = 3.2 M Solutions with a [H+] higher than 1.0 M have a negative pH.
8. The Problem 7 solution has the both the lowest pH and the highest [H+]. By either measure it is the most
acidic. Practice D: pOH Calculations
1. If [OH─] = 10─5,
2. If [H+] = 10─4, a. [H+] = 10─9 M b. pH = 9 c. pOH = 5 d. Basic a. [OH─] = 10─10 M b. pH = 4 c. pOH = 10 d. Acidic 3. If pH = 12 , a. [H+] = 10─12 M
4. If pOH = 8 a. pH = 6 © 2011 ChemReview.Net v. 2m b. [OH─] = 10─2 M b. [H+] = 10─6 M c. pOH = 2 d. Basic c. [OH─] = 10─8 M d. Acidic Page 936 Module 29 — AcidBase Fundamentals 5. a. WANT: ? mol NaOH = [NaOH]mixed
L soln
DATA: 2.00 grams NaOH = 750. mL soln (equivalent: two measures of same solution.) 40.0 grams NaOH = 1 mole NaOH (grams prompt) SOLVE: ? moles NaOH = 2.00 g NaOH ● 1 mol NaOH ● 1 mL
liter of soln
750 mL soln
40.0 g NaOH 10─3 L = 0.0667 M NaOH b. See NaOH and [ions]? Write the REC steps: Reaction, Extent, Concentrations.
R and E:
C: 1 NaOH
^
0.0667 M 0 M 1 Na+ + 1 OH─ (~100%)
^
^
0.0667 M
0.0667 M [NaOH]mixed = 0.0667 M = [OH─]in soln. = [Na+] in soln.
b. [OH─] = ? [NaOH] mxd. = 0.0667 M (part a) = [OH─]in soln = [Na+]in soln = Part c answer
d. [H+] WANT [H+], know [OH─] .
[H+] = 1.0 x 10─14 = 1.0 x 10─14 = 0.15 x 10─12 = 1.5 x 10─13 M H+
[OH─]
6 .67 x 10─2
e. pOH = ?
See pOH? Write: pOH ≡ ─ log[OH─] and [OH─] ≡ 10─pOH and pH + pOH = 14.00
pOH ≡ ─ log[OH─] = ─ log(0.0667) = ─ log (6.67 x 10─2) = 1.18 = pOH
f. pH? Since pH + pOH = 14.00 , pH = 14.00 – pOH = 14.00 ─ 1.18 = 12.82 (compare pH to [H+] above.) 6. The pOH in a KOH solution is 1.5 . What is the [KOH]?
b. See KOH and [particles]? Write the REC steps. See pOH? Use the pOH prompt.
R and E:
C: 1 KOH
^
XM 0M [OH─] ≡ 10─pOH 1 K+
^
XM + 1 OH─
^
XM = 10─1.5 M = 3 x 10─2 M (~100%) = [OH─] = X M = [KOH]mixed ***** © 2011 ChemReview.Net v. 2m Page 937 Module 29 — AcidBase Fundamentals Summary  AcidBase Fundamentals
If you have not already done so, you may want to add these rules to your flashcards. AcidBase Fundamentals
1. See [H+] and [OH─] ? Write: Kw = [H+][OH─] = 1.0 x 10─14
2. In water,
• strong monoprotic acids ionize ~100% to form H+. • Alkali metal hydroxides ionize 100% to form OH─. • For [ions], write the REC steps or use the quick rules. 3. Quick rules: [HCl or HNO3]mixed = [H+]in soln.
[NaOH or KOH]mixed = [OH─]in soln. 4. See [H3O+] ? Write: [H3O+] = [H+]
5. In acid solutions, use acid ionization rules to find [H+], then Kw to find [OH─].
6. In base solutions, use base rules to find [OH─], then Kw to find [H+].
7. See pH ? Write: pH ≡ ─ log [H+] and [H+] = 10─pH 8. See pOH ? Write: pOH ≡ ─ log[OH─] and [OH─] = 10─pOH
and pH + pOH = 14.00 (at 25ºC) 8. pH values are not given units. When a concentration is calculated based on pH, the
unit mol/L (M) must be assigned to the answer.
9.. Using the pH scale, for aqueous solutions at 25ºC:
a. Pure water has a pH of 7.0
b. In a neutral solution, [H+] = [OH─] = 1.0 x 10─7 M; and pH = 7.
c. In acidic solutions, pH is less than 7. In basic solutions, pH is greater than 7.
d. A lower pH means a higher acidity. A higher pH means a higher basicity.
e. The further from 7 is pH, the stronger is the acidity or basicity of the solution. ##### © 2011 ChemReview.Net v. 2m Page 938 ...
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