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General Chemistry/Acids and Bases/Determining pH and pOH

Acids and Bases

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Determining pH and pOH

A measure of the concentration of hydronium ions (H3O+) in a solution is pH, and a measure of the concentration of hydroxide ions is pOH. The pH scale ranges from 1 to 14, with 1 being most acidic and 14 being most basic.

The pH of a substance is a measure of the concentration of H+ (or H3O+) ions in solution. The relationship is logarithmic (base 10): the pH value itself is the value of the exponent, x, when the concentration is expressed as 1×10−xM1\times{10}^{-x}{\;\rm{M}}. In other words, in a substance that has a pH of 5, the concentration of H+ ions is 1×10−5M1\times{10}^{-5}\;{\rm{M}}. The pOH of a substance is a measure of the concentration of OH– ions in solution. The total of the exponents must equal 14, so in a substance with a pH of 5, the pOH must be 9. When the two values are compared, 1×10−5M1\times{10}^{-5}\;{\rm{M}} and 1×10−9M1\times{10}^{-9}\;{\rm{ M}}, the concentration of H+ is greater than that of OH– by four orders of magnitude. When [H+] is greater than [OH–], the solution is acidic. When the opposite is true, the solution is basic.

The following equation relates pH to the concentration of H+:
pH=−log⁡[H+]\rm{pH}=-\log{[\rm {H}^{+}]}
For example, calculating [H3O+] of 0.20 M acetic acid (CH3COOH) at equilibrium gives a concentration of 1.9×10−31.9\times10^{-3}. Substitute x for [H+] and [CH3COO–] in the equation for the equilibrium constant, and substitute 0.20 for [CH3COOH]:
Ka=[H+][A−][HA]=x20.20K_{\rm {a}}=\frac{\lbrack\rm {H}^{+}\rbrack\lbrack\rm {A}^{-}\rbrack}{\lbrack\rm{HA}\rbrack}=\frac{x^2}{0.20}
Substitute the known value of Ka, 1.77×10−51.77\times10^{-5}, and rearrange the equation to solve for x:
x2=(1.77×10−5)(0.20)x2=3.54×10−6x=3.54×10−6=1.9×10−3\begin{aligned}x^2&=(1.77\times10^{-5})(0.20)\\x^2&=3.54\times10^{-6}\\x&=\sqrt{3.54\times10^{-6}}\\&=1.9\times10^{-3}\end{aligned}
The value of x equals the concentration of the hydronium ion.
x=1.9×10−3=[H+]x=1.9\times10^{-3}=\lbrack{\rm {H}^{+}}\rbrack
The negative logarithm is the pH of the solution.
pH=−log⁡(1.9×10−3)=2.7{\rm{pH}}=-\log(1.9\times10^{-3})=2.7
Thus the pH can be calculated knowing only Ka and the concentration of an acid.
Because pH and pOH must add up to 14, the pOH of the solution can also be determined:
pOH=14−pH=14−2.7=11.3\begin{aligned}\rm{pOH}&=14-\rm{pH}\\&=14-2.7\\&=11.3\end{aligned}
The values of pH and pOH can also be used to calculate [H3O+] or [OH–].
Step-By-Step Example
Using pH and pOH to Calculate [H3O+] and [OH–]
What is [H3O+] of a solution with a pOH of 8.3?
Step 1
First, determine the pH by subtracting pOH from 14.
pH=14−pOH=14−8.3=5.7\begin{aligned}\rm{pH}&=14-\rm{pOH}\\&=14-8.3\\&=5.7\end{aligned}
Step 2
Substitute this value into the pH equation.
pH=−log⁡[H3O+]5.7=−log⁡[H3O+]\begin{aligned}\rm{pH}&=-\log[{\rm{H}_{3}\rm{O}^{+}}]\\5.7&=-\log[{\rm{H}_{3}\rm{O}^{+}}]\end{aligned}
Solution
Solve for [H3O+].
[H3O+]=10−5.7=2.0×10−6\begin{aligned}\left[{\rm {H}_{3} O^{+}}\right]&=10^{-5.7}\\&=2.0\times10^{-6}\end{aligned}

pH Scale

Familiar substances have a range of pH values.
In some situations it is desirable to limit the pH change that occurs when an acid or a base is added to a solution. Acidity changes in bodily fluids, for example, must remain within a certain range to avoid health issues. A buffer is a solution that contains significant quantities of an acid and its conjugate base and resists a change in pH when acid or base is added. The pH of a buffer system depends on Ka for the weak acid as well as the initial concentrations of the acid [HA] and the base [A–] that are mixed. The Henderson-Hasselbalch equation can be used to approximate the pH of a buffer solution:
pH≈pKa+log⁡10[A−][HA]{\rm{pH}}\approx\rm{p}K_{\rm{a}}+\log_{10}\frac{\rm{[A}^{-}]}{[\rm{HA}]}
The term pKa in the equation is defined in terms of the acid ionization constant:
pKa=−log⁡Ka{\rm{pK}_{a}}=-\log{K}_{\rm{a}}
Step-By-Step Example
Using the Henderson-Hasselbalch Equation to Calculate the pH of a Buffer
Consider a buffer solution of 0.035 M NH3 and 0.050 M NH4+. NH4+ has a Ka of 5.6×10−105.6\times10^{-10}. What is the pH of the buffer?
Step 1
Use the acid ionization constant to calculate pKa.
pKa=−log⁡Ka=−log⁡(5.6×10−10)=9.25\begin{aligned}{\rm{p}K_{\rm{a}}}&=-\log{K}_{\rm{a}}\\&=-\log({5.6\times10^{-10}})\\&=9.25\end{aligned}
Solution
Substitute pKa and the molarities in the Henderson-Hasselbalch equation to calculate the pH.
pH≈pKa+log⁡10[A−][HA]≈pKa+log⁡10[NH3][NH4+]≈9.25+log⁡10[0.035][0.050]≈9.1\begin{aligned}\rm{pH}&\approx\rm{p}K_{\rm{a}}+\log_{10}\frac{\rm{[A^{-}]}}{[\rm{HA}]}\\&\approx\rm{pK_{a}}+\log_{10}\frac{\rm{[NH}_{3}]}{[{\rm{NH}_{4}}^{+}]}\\&\approx9.25+\log_{10}\frac{[0.035]}{[0.050]}\\&\approx9.1\end{aligned}
The pH of the buffer is 9.1.
The Henderson-Hasselbalch equation can be rearranged to give the ratio of the concentrations of the base [A−] and the acid [HA].
10pH−pKa=[A−][HA]10^{\rm{pH-pK}_{\rm{a}}}=\frac{\rm{[A^{-}]}}{[\rm{HA}]}
This equation can be used to calculate the amount of acid and conjugate base needed to make a buffer of a particular pH.
<Strengths of Acids and Bases>Suggested Reading