Question 7.16: The pH of Salts Calculate the pH of a 0.100 M solution of NH...

The major species in solution are

The familiar reactions involving these species are

NH_{4} ^{+} (aq ) \xrightleftharpoons[]{}  NH_{3} (aq)  + H^{+} (aq)                                                                  K_{a} = 5.6  × 10 ^{-10}

CN^{-} (aq ) + H_{2}O (l)\xrightleftharpoons[]{}  HCN (aq)  + OH^{-} (aq)                            K_{b} = 1.6  × 10 ^{-5}

H_{2}O (l) + H_{2}O (l)\xrightleftharpoons[]{}  H_{3}O^{+} (aq)  + OH^{-} (aq)                            K_{w} = 1.0  × 10 ^{-14}

However, noting that NH_{4} ^{+} is an acid and CN^{-} is a base, it is also sensible to consider the reaction

NH_{4} ^{+} (aq) +  CN^{-}(aq ) \xrightleftharpoons[]{}  HCN (aq)  + NH_{3} (aq) 

To evaluate the importance of this reaction, we need the value of its equilibrium constant. Note that

K= \frac{[NH_{3}]  [HCN]}{[NH_{4} ^{+}] [CN^{-} ]} = \frac{[H^{+}] [NH_{3}]}{[NH_{4} ^{+}]}  × \frac{[HCN]}{[H^{+}] [CN^{-} ]}

= K _{a} (NH_{4} ^{+} ) × \frac{1}{K _{a} (HCN)}  

Thus the value of the equilibrium constant we need is

K= \frac{K _{a} (NH_{4} ^{+} )}{K _{a} (HCN)}  = \frac{5.6  × 10 ^{-10}}{6.2  × 10 ^{-10}} = 0.90

Notice that this equilibrium constant is much larger than those for the other possible reactions. Thus we expect this reaction to be dominant in this solution.
Following the usual procedures, we have the concentrations listed below.

Then                                    K= 0.90 = \frac{x²}{(0.100- x)²}

Taking the square root of both sides yields

0.95 = \frac{x}{0.100 – x}

and

x= 4.9 × 10 ^{-2}    M    =  [NH_{3}]  = [HCN] 

Notice that the reaction under consideration does not involve H^{+} or OH^{-} directly. Thus, to obtain the pH, we must consider the position of the HCN or NH_{4} ^{+}   dissociation equilibrium. For example, for HCN

  K_{a} = 6.2  × 10 ^{-10}  = \frac{[H^{+}] [CN^{-}]}{[HCN]}

From the preceding calculations,

[CN^{-}] = 0.100  –  x =  0.100  –  0.049  = 0.051   M 

[HCN] = x  =  4.9 × 10 ^{-2}  M

Substituting these values into the K_{a} expression for HCN gives

[H^{+} ] = 6.0  × 10 ^{-10}   M

and

pH =9.22

Note that this solution is basic, just as we predicted in Example 7.15.