Acids and Bases 2

Acids and Bases 2 - Salt Hydrolysis, Polyprotic Acids and...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Salt Hydrolysis, Polyprotic Acids and Indicators
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
02/27/08 2 Consider the situation where solutions of salts of weak acids and bases exhibit pH values different from 7 because of reaction of one of the ions with the solvent, leaving a net excess of either [H 2 O + ] or [OH ].
Background image of page 2
02/27/08 3 Example: Calculate the pH of 0.1 M sodium acetate (CH 3 CO 2 Na) in H 2 O First, we assume that the sodium salt of the weak acid is completely dissociated: CH 3 CO 2 Na (aq) CH 3 CO 2 (aq) + Na + (aq) (As a rule, one may assume that all compounds of alkali metals are highly soluble in water, dissociating to the component ions. The same applies to most ammonium salts and most nitrates). For this reason, I draw the reaction with a single arrow to indicate a complete reaction.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
02/27/08 4 With the above assumption, the problem reduces to the hydrolysis of the acetate ion. We take this to be the conjugate base of acetic acid, so that the equilibrium constant for this reaction is defined as K b . CH 3 CO 2 (aq) + H 2 O CH 3 CO 2 H (aq) + OH (aq) There are two ways to obtain a value for the equilibrium constant. First, we can write down the equilibrium constant, multiply top and bottom by [H 3 O + (aq)], and then rearrange, as follows: K b = [ CH 3 CO 2 H ][ OH ] [ CH 3 CO 2 ] = [ CH 3 CO 2 H ][ OH ][ H 3 O ] [ CH 3 CO 2 ][ H 3 O ] = K W K a
Background image of page 4
02/27/08 5 Instead of this ratio, we could use logarithmic notation, specifically: -log 10 K b = -log 10 (K w /K a ) = -log 10 K w - (-log 10 K a ) or simply: pK b = pK w - pK a and, since pK w = 14 at 298 K: pK a + pK b = 14 This is true for any acid-conjugate base pair.
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
02/27/08 6 Alternatively, we could synthesize the reaction from the following: CH 3 CO 2 - (aq) + H 3 O + (aq) CH 3 CO 2 H (aq) + H 2 O (1) 2H 2 O H 3 O + (aq) + OH (aq) (2) Here, K 1 = K a -1 and K 2 = K w . Adding the two equations gives the equation whose equilibrium constant K b = K w /K a . CH 3 CO 2 - (aq) + H 2 O CH 3 CO 2 H (aq) + OH (aq) (3) Therefore, for acetic acid, we are given that K a = 1.8 x 10 -5 , so that K b for the acetate ion has the value 10 -14 / (1.8 x 10 -5 ) = 5.6 x 10 -10 .
Background image of page 6
02/27/08 7 Now, to solve the sodium acetate problem, we write as usual:
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/29/2008 for the course HIST 101 taught by Professor Wormer during the Spring '08 term at UPenn.

Page1 / 23

Acids and Bases 2 - Salt Hydrolysis, Polyprotic Acids and...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online