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 1

Solution :-

Reaction equation

Ag+ + 2CN-   --------- > Ag(CN)2-

Kf= 3*10^-20

Lets first calculate the moles of the Ag+ and CN-

Moles of Ag+ = molarity * volume

                         = 2.6*10^-3 mol per L * 0.130 L

                         = 3.38*10^-4 mol Ag+

Moles of CN- = 0.130 mol per L * 0.220 L = 0.0286 mol CN-

Mole ratio of the Ag+ to CN- is 1 : 2

So lets calculate the moles of the CN- needed to react with Ag+ moles

3.38*10^-4 mol Ag+ * 2 mol CN- / 1 mol Ag+ = 6.76*10^-4 mol CN-

Now lets find the moles of the CN- remain after the reaction

Moles of CN- remain = 0.0286 mol – 6.76*10^-4 mol = 0.027924 mol CN-

Now lets calculate the new molarioty of the Ag(CN)2- and CN- at total volume

Total volume = 130 ml + 220 ml = 350 ml = 0.350 L

[Ag(CN)2-] = 0.00338 mol / 0.350 L = 0.009657 M

[CN-] = 0.027924 mol / 0.350 L = 0.079783 M

Using the Kf lets calculate the concentration of the [Ag+]

Kf= [Ag(CN)2-] /[Ag+][CN-]^2

3.0*10^20 = [0.009657] / [Ag+] [ 0.079783]^2

[Ag+] = ([0.009657] / 3.0*10^20) / [ 0.079783]^2

[Ag+] = 5.06*10^-21 M

So the concentration of the Ag+ = 5.06*10^-21 M