Question

The solubility of silver chloride can be increased by dissolving it in a solution containing ammonia. agcl (s) ag+ (aq) + cl- (aq) k1 = 1.6 x 10-10 ag+ (aq) + 2nh3 (aq) ag(nh3)2+ (aq) k2 = 1.5 x 107 what is the value of the equilibrium constant for the overall reaction? agcl (s) + 2nh3 (aq) ag(nh3)2+ (aq) + cl- (aq) knet = ?

Answer 1

The correct equation for the overall reaction can simply be obtained by adding the two separate equations together. Now when you add the two equations together, the overall K can be calculated by multiplying the individual K values. Therefore:

K(overall) = K1 * K2

K(overall) = (1.6 x 10^-10) * (1.5 x 10^7)

K(overall) = 2.4 x 10^-3

Answer 2

Answer:

$K_{net}=[Cl^-]*[Ag(NH_3)_2^{+2}]=2.4*10^{-3}$

Explanation:

Silver chloride dissosation equation:

$AgCl \longrightarrow Ag^+ + Cl^-$

$K_1=[Ag^+]*[Cl^-]$

Reaction with ammonia:

$Ag^+ + 2 NH_3 \longrightarrow Ag(NH_3)_2^{+2}$

$K_2=\frac{[Ag(NH_3)_2^{+2}]}{[Ag^+]}$

Overall reaction:

$AgCl + 2 NH_3 \longrightarrow Ag(NH_3)_2^{+2} + Cl^-$

$K_{net}=K_1 * K_2$

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

$K_{net}=[Cl^-]*[Ag(NH_3)_2^{+2}]$

$K_{net}=1.6*10^{-10}*1.5*10^7=2.4*10^{-3}$