$B A+H_{2} O \rightleftharpoons B O H+H A$
Now $p H$ is given by
$p H=\frac{1}{2} p K_{w}+\frac{1}{2} p K a-\frac{1}{2} p K_{b}$
Substituting given values, we get
$p H=\frac{1}{2}(14+4.80-4.78)=7.01$
$K_{sp}$ એ $AgCl = 1.2\times 10^{-10} \,K_{sp}$ એ $AgI = 1.7 \times 10^{-16}$
$K_{sp}$ એ $AgSCN = 7.1 \times 10^{-7} \,K_{sp}$ એ $AgBr = 3.5 \times 10^{-13}$