Question

Determine the concentration of nh3(aq) that is required to dissolve 743 mg of agcl(s) in 100.0 ml of solution. the ksp of agcl is 1.77Ã 10â10.

Answer 1

Answer:

1.1 M

Explanation:

The dissociation of $AgCl_{(s)}$ is as follows:

$AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}$

Given Value for $K_{sp} = 1.77*10^{-10}$

The equation for the reaction for the formation of complex ion $Ag(NH_3)^+_2$ is :

$Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}$

The value of $K_f = 1.6*10^7$

If we combine both equation and find the overall equilibrium constant will be:

$AgCl \leftrightharpoons Ag^+_{(aq)}+Cl^-_{(aq)}$

$Ag^+_{(aq)} + 2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)}$

                                                                                                     

$AgCl_{(s)}+2NH_3_{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}$

                                 $K = (1.77*10^{-10})(1.6*10^7)$

                                  $K = 0.00283$

If $[NH_3]$ = x M

The solubility of  $AgCl_{(s)}$ in the $NH_3$ solution will be:

$x = 743*10^{-3}g * \frac{mol AgCl}{143.32g}*\frac{1}{0.1000L}$

$x = 0.0518 M$

Constructing an ICE Table; we have :

                             $AgCl_{(s)} + 2 NH_3{(aq)} \rightleftharpoons Ag(NH_3)_2^+_{(aq)} + Cl^-_{(aq)}$

Initial  (M)                                     x                   0                      0

Change  (M)                       -2 (0.0518)      + 0.0518          + 0.0518

Equilibrium (M)                    x - 0.1156          0.0518             0.0518

Equilibrium constant;

(K) = $\frac{[Ag(NH_3)_2^+][Cl^-]}{[NH_3]^2}$

$0.00283 = \frac{(0.0518)^2}{(x-0.1156)^2}$

$0.00283 = (\frac{0.0518}{x-0.1156})^2$

$x = 0.1156 + \sqrt{\frac{0.0518^2}{0.00283} }$

x = [NH₃] =  1.089 M

[NH₃] ≅ 1.1 M