Acids and Bases 2

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

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Salt Hydrolysis, Polyprotic Acids and Indicators

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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 ].
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.

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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
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.

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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 .
02/27/08 7 Now, to solve the sodium acetate problem, we write as usual:

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

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

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