Question

$ {H_3}P{O_4} $ is a weak triprotic acid: approximate $ pH $ of $ 0.1M $ $ N{a_2}HP{O_4}(aq) $ is calculated by:
(A) $ \dfrac{1}{2}[p{K_{a1}} + p{K_{a2}}] $
(B) $ \dfrac{1}{2}[p{K_{a2}} + p{K_{a3}}] $
(C) $ \dfrac{1}{2}[p{K_{a1}} + p{K_{a3}}] $
(D) $ \dfrac{1}{2}[p{K_{a1}} + p{K_{a2}}] $

Answer 1

Hint: Triprotic acid can donate three hydrogen ions per molecule during dissociation. Stepwise dissociation constants are each defined for the loss of a single proton. The constant for dissociation of the first proton may be denoted as $ {K_{a1}} $ and the constants for dissociation of successive protons as $ {K_{a2}} $ , etc.

Complete answer:
The $ p{K_a} $ value is one method used to indicate the strength of an acid. $ p{K_a} $ is the negative log of the acid dissociation constant . A lower $ p{K_a} $ value indicates a stronger acid. That is, the lower value indicates the acid more fully dissociates in water.
 $ \Rightarrow {K_a} = \dfrac{{[{A^ - }][{H^ + }]}}{{[HA]}} $
Where, $ {K_a} = $ acid dissociation constant
 $ \Rightarrow [{A^ - }] = $ Concentration of the conjugate base of the acid
 $ \Rightarrow [{H^ + }] = $ Concentration of hydrogen ions
 $ \Rightarrow [HA] = $ Concentration of chemical species $ HA $
 $ 2HP{O_4}^{2 - } \leftrightarrow P{O_4}^{3 - } + {H_2}P{O_4}^ - $
From this we see that, $ [{H_2}P{O_4}^ - ] \cong [P{O_4}^{3 - }] $
 $ \Rightarrow {K_{a2}} = \dfrac{{[{H_3}{O^ + }][HP{O_4}^{2 - }]}}{{[{H_2}P{O_4}^ - ]}} $
 $ \Rightarrow {K_{a3}} = \dfrac{{[{H_3}{O^ + }][P{O_4}^{3 - }]}}{{[HP{O_4}^{2 - }]}} $
When, $ [{H_2}P{O_4}^ - ] = [P{O_4}^{3 - }] $
 $ \Rightarrow \dfrac{{[{H_3}{O^ + }][HP{O_4}^{2 - }]}}{{{K_{a2}}}} $ $ $
 $ \Rightarrow {K_{a3}}\dfrac{{[{H_3}{O^ + }][P{O_4}^{3 - }]}}{{[HP{O_4}^{2 - }]}} $
 $ \Rightarrow {[{H_3}{O^ + }]^2} $
Taking log both sides,
 $ \Rightarrow \log {K_{a3}} + \log {K_{a2}} = 2\log [{H_3}{O^ + }] $
Or, $ pH = \dfrac{1}{2}(p{K_{a2}} + p{K_{a3}}) $
So, the correct answer is B) $ pH = \dfrac{1}{2}(p{K_{a2}} + p{K_{a3}}) $ .

Additional Information:
A knowledge of $ p{K_a} $ values is important for the quantitative treatment of systems involving acid–base equilibria in solution. Many applications exist in biochemistry. for example, the $ p{K_a} $ values of proteins and amino acid side chains are of major importance for the activity of enzymes and the stability of proteins.

Note:
When the difference between successive $ p{K_a} $ values is about four or more, as in this example, each species may be considered as an acid in its own right. In fact salts of $ {H_2}P{O_4}^ - $ may be crystallized from solution by adjustment of $ pH $ to about $ 5.5 $ and salts of $ HP{O_4}^ - $ may be crystallized from solution by adjustment of $ pH $ to about $ 10 $ .