$\left[\right.$ આપેલ છે: ${K}_{m}=1 \times 10^{-14}$ અને $\left.{K}_{{b}}=1.8 \times 10^{-5}\right]$
So \({K}_{b}=\frac{\left[{NH}_{4}^{+}\right]\left[{HO}^{-}\right]}{\left[{NH}_{3}\right]}\)
\(\left[{HO}^{-}\right]=\frac{{K}_{{b}} \times\left[{NH}_{3}\right]}{\left[{NH}_{4}^{+}\right]} =1.8 \times 10^{-5} \times \frac{2}{5} \times \frac{210}{504}\)
\(=3 \times 10^{-6}\)
$A = NH_4Cl$; $ B = CH_3COONa$; $ C = NH_4OH$; $D = CH_3COOH$
$AgCl\downarrow +2N{{H}_{3}}\rightleftharpoons {{\left[ Ag{{\left( N{{H}_{3}} \right)}_{2}} \right]}^{+}}+C{{l}^{-}}$
નીચેના પૈકી શુ ઉમેરવાથી $AgCl$ ની સફેદ અવક્ષેપ મળશે ?