The `pK_(a)` of `CH_(3)COOH` and `pK_(a)` of `nH_(4)OH` is `4.76` and `4.75`, respectively. Calculate the hydrolysis constant of ammonium acetate `(CH_(3)COONH_(4))` at `298K` and also the drgree of hydrolysis and `pH` of its (a) `0.01M` and (b) `0.04M` solutions.

Explanation

(i)`pK_a`=7.76
-log `K_a` = 4.76
or `log K_a`=-4.76 =5.24
`therefore K_a`=antilog (5.24) = `1.74xx10^(-5)`
`pK_b` =4.75
-log `K_b` =4.75 or log `K_b`=-4.75 =5.25
`therefore K_b` =antilog (5.25) =`1.78xx10^(-5)`
Now `K_h=K_w/(K_axxK_b)=10^(-14)/((K_axxK_b)xx(1.78xx10^(-5)))`
`=3.23xx10^(-5)`
(ii) Since `K_h` is very small
`h=(K_h)^(1/2) =(3.23xx10^(-5))^(1/2)`
`therefore h=5.68xx10^(-3)`
pH of solution of salt of weak acid and weak base if
(iii)`pH=7+1/2(pK_a+pK_b)`
`=7+1/2(4.76-4.75)`
=7.005
Since pH is independent of concentration, it is same for (a) and (b).