\(\quad \quad \quad 2.2 \times 10^{-4} \mathrm{M} \quad 1.1 \times 10^{-4} \mathrm{M}\)
\(\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]\)
\(=\left[2.2 \times 10^{-4}\right]^{2} \cdot\left[1.1 \times 10^{-4}\right]\)
\(\mathrm{K}_{\mathrm{sp}}=5.3 \times 10^{-12}\)
[અહીં $K_b\,(NH_4OH) = 10^{-5}$ અને $log\,2 = 0.301$ ]