Calculate the solubility of AgBr in 1.00 M aqueous ammonia. K_sp(AgBr) = 5.0x 10^-13; K_f of [Ag(NH3)2]+ = 1.6x 10^7. Treat [Ag(NH3)2]+ as the only dissolved silver species and take the free ammonia as 1.00 - 2s. Give s in units of 10^-3 M to two decimals.

JEE Main levelNumerical answerDifficulty 2 of 3ia155

Calculate the solubility of AgBr in $1.00 M$ aqueous ammonia. $K_{s p} (AgBr) = 5.0 \times 1 0^{- 13}$ ; $K_{f}$ of $[Ag (NH X_{3}) X_{2}] X^{+} = 1.6 \times 1 0^{7}$ . Treat $[Ag (NH X_{3}) X_{2}] X^{+}$ as the only dissolved silver species and take the free ammonia as $1.00 - 2 s$ . Give $s$ in units of $1 0^{- 3} M$ to two decimals.

Numerical answer type, answer in x 10^-3 M, tolerance 0.02.

K = K_{s p} K_{f} = 8.0 \times 1 0^{- 6}

for

AgBr + 2 NH X_{3} [Ag (NH X_{3}) X_{2}] X^{+} + Br X^{-}

s / (1.00 - 2 s) = K = 0.00283

, hence

s = 0.00283/ (1 + 2 \times 0.00283) = 2.8125 e - 03 M

, i.e.

2.81 \times 1 0^{- 3} M