Question

What is [I−] when 125 mL of 0.0010 M KI is reacted with 75 mL of 0.0010 M HgCl2? 4 KI(aq) + HgCl2(aq) --> 2 KCl(aq) + K2HgI4(aq) Kf (HgI4) = 2 x 1030

Answer 1

4 KI(aq) + HgCl2(aq) --> 2 KCl(aq) + K2HgI4(aq)           Kf (HgI4) = 2 x 1030

kf = [KCl]^2[K2HgI4]/[HgCl2][KI]^4

concentration of KI in the solution = 0.001*125/200 = 0.000625 M

concentration of HgCl2 in the solution = 0.001*75/200 = 0.000375 M

2*10^30 = (x^2*x)/((0.000625)^4*(0.000375))

at equilbrium,

[KCl] = x = 4.85*10^4 M

[K2HgI4] = x = 4.85*10^4 M

[HgCl2] = 0.000625-4x = 0.000625 M

[KI] = 0.000375 - x = 0.000375 M

so that,

[I-] = [KI] = 0.000375 M