Question

A 125.0 −mL sample of a solution that is 2.8×10⁻³ M in AgNO3 is mixed with a 220.0 −mL sample of a solution that is 0.10 M in NaCN . After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

I have tried different ways and come up with these answers, all incorrect:

[Ag+] = 5.5e-15 M

[Ag+] = 1.9e-15 M

[Ag+] = 1.3e-22 M

Please show work.

Answer 1

Here in this problem we show the reaction between NaCN and AgNO3.

By using molarity and volume we get moles of both. Once we get moles of both reactants then we can get limiting reactant. We find excess moles of excess reactants. From these moles we get equilibrium moles of Ag+ and then concentration.

Reaction:

AgNO3 (aq) + NaCN (aq) --- > Ag(CN) (s) + NaNO3 (aq)

Moles of Ag(NO3) = Volume in L x molarity = 0.125 L x 1.28 E-3 M = 0.00016 mol

Moles of NaCN = 0.220 L x 0.10 M = 0.0220 mol

Since from reaction mole ratio of these two reactions = 1:1

So the moles of Ag(NO)3 are less and so it is limiting reactant.

Limiting reactant is consumed completely in the reaction. Excess reactant is NaCN.

Since limiting reactant contains Ag+ and so there will be any free Ag+ so we have to use ksp value of AgCN in order to get Ag+ concentration.

Ksp of AgCN = 1.6 E-14

From this ksp we calculated solubility of Ag+

AgCN (s) --- > Ag+ + CN-

Both the ions are in same ration and in this situation we calculate solubility by using following equation.

Ksp= x²

so x = sqrt (1.6 E-14)

= 1.265 E-7 M

This also means that moles of AgCN produces moles of Ag+ in in 1 L = 1.265 E-7

Now we calculate number of moles of AgCN formed

Moles of AgCN formed in the reaction = Number of moles of AgNO3 x 1 mol AgCN/ 1 mol AgNO3

We used AgNO3 because it is limiting reactant.

So the number of moles of AgCN formed = 0.00016 mol

Number of moles of Ag + produced

= 0.00016 mol x (1.265 E-7/ 1 mol)

= 2.02 E -11 mol

Lets calculated molarity of it.

[Ag+] = 2.0 E-11 mol / volume (total ) in L

[Ag+] = 2.0 E-11 mol /(0.125+0.220) L

= 5.87 E-11 M