Question

A 115.0 −mL sample of a solution that is 2.6×10−3 M in AgNO3 is mixed with a 230.0 −mL sample of a solution that is 0.14 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

moles of AgNO3 = 2.6 x 10^-3 x 115 /1000 = 3 x 10^-4 = 0.0003

moles of NaCN = 0.14 x 230 / 1000 = 0.0322

AgNO3 + 2 NaCN -----------------------------------> [Ag(CN)2 ]-    + NaNO3

0.0003       0.0322                                                  0.0003                  0.0003

Ag+    + 2 CN- <------------------------> Ag(CN)2-

Kf = 0.0003 / [Ag+] [0.0316]^2

1.0x 10^21 = 0.0003 / [Ag+] [0.0316]^2

[Ag+] = 3 x 10^-22 M