Question

A 115.0 −mL sample of a solution that is 2.9×10⁻³ M in AgNO3 is mixed with a 230.0 −mL sample of a solution that is 0.10 M in NaCN.

After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Express your answer using two significant figures.

Answer 1

Moles of Ag⁺ = 0.115 L x 2.9 x 10⁻³ M / L

= 0.0003335 moles of Ag⁺

Moles of CN^- = 0.230 L x 0.10M / L

= 0.023 moles of CN^-

we can say that , when 0.0003335 moles of Ag⁺ added with 0.023 moles of CN^- , all moles of Ag⁺ converted to [Ag(CN)₂]^-

total volume = 0.115 + 0.230 = 0.345 L

so

0.0003335 moles of [Ag(CN)₂]^- / 0.345L = 0.000967 M [Ag(CN)₂]^-

Also

1 Ag⁺ and 2 CN^- [Ag(CN)₂]^-

so twice 0f CN- consumed when it reacts with 0.0003335 moles of Ag⁺

so

(0.023 moles of CN^- ) - 2 x ( 0.0003335 moles of Ag⁺ ) = 0.02233 moles of CN^- remains

and this gives (0.02233 moles of CN^-) / 0.345 L = 0.0647 M CN^-

Now

Kf = [Ag⁺(CN^-)2] / [Ag⁺][CN^-]²

1 x 10²¹ = 0.000967 M / ( [Ag⁺] x (0.0647 M)²)

[Ag⁺] = 0.231 x 10^-21 M

= 2.31 x 10^-22 M