Question

A 130.0 −mL sample of a solution that is 2.6×10−3 M in AgNO3 is mixed with a 220.0 −mL sample of a solution that is 0.13 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains? Answer with 2 sig. figures.

Answer 1

V = 130 ml

M = 2.6*!0^-3 AgNO3

V = 220 ml

M = 0.13 NaCN

this will form the next equilibirum

AgCN

all other Ag+ will be precipitated as AgCn

only Ag+ in equilibrium will remain

AgCN <-> Ag+ + CN-

Ksp = [Ag+][CN-]

The value of KSp at 25°C is

Ksp = 1.2*10^-16

Then

calculate mols of each

mol of Ag+ initial = M*V = 130*(2.6*10^-3) = 0.338 mmol of Ag+

mol of Cn- initial MV = 0.13 *220 = 28.6 mmol of CN-

CN- is clearly limiting so

Ksp = [Ag+][CN-]

Ksp = S*S

1.2*10^-6 = S^2

S = sqrt(1.2*10^-6) = 0.00109544511

2 sig fi

1.1*10^-3 mol per liter