At 298 K, solid AgCl (Ksp=1.8×10^-10)…

At 298 K, solid AgCl (Ksp=1.8×10^-10) is shaken with 0.10 M aqueous NH3, forming [Ag(NH3)2]^+ with formation constant Kf=1.6×10^7. Assuming that all…

Numerical answer

298\ \mathrm{K}

, solid

\mathrm{AgCl}

(

K_{sp}=1.8\times10^{-10}

) is shaken with

0.10\ \mathrm{M}

aqueous

\mathrm{NH_3}

, forming

[\mathrm{Ag(NH_3)_2}]^+

with formation constant

K_f=1.6\times10^{7}

. Assuming that all dissolved silver is present as the complex and that the ammonia consumed is

2s

, where

s

is the molar solubility of

\mathrm{AgCl}

, the value of

s\times10^{3}

(in

\mathrm{mol\,L^{-1}}

scaled by

10^3

) is ____. Give the answer correct to two decimal places.

Show answer and solution

Answer: 4.85

The overall reaction is

\mathrm{AgCl(s)+2NH_3\rightleftharpoons[Ag(NH_3)_2]^++Cl^-}

with

K=K_{sp}K_f=1.8\times10^{-10}\times1.6\times10^{7}=2.88\times10^{-3}

. If

s

mol L

^{-1}

dissolves,

[\mathrm{Ag(NH_3)_2^+}]=[\mathrm{Cl^-}]=s

and

[\mathrm{NH_3}]=0.10-2s

. Therefore

\frac{s^2}{(0.10-2s)^2}=2.88\times10^{-3}

, so

\frac{s}{0.10-2s}=\sqrt{2.88\times10^{-3}}=0.05367

. Solving,

s=0.005367-0.10734\,s

, so

s(1.10734)=0.005367

and

s=4.846\times10^{-3}\ \mathrm{M}

. Hence

s\times10^3\approx4.85

. Ignoring the ammonia consumed would give

5.37

, which is the common error.

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