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 0.10 M in NaCN.

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

Answer 1

115.0 mL sample of a solution that is 2.6×10^-3 M in AgNO3 is
(0.115 L ) (2.6×10−3 mol / L) = 0.000299 moles of Ag+

230.0 mL sample of a solution that is 0.10 M in NaCN is
(0.230 L) (0.10 M in NaCN) = 0.023 moles of CN-

when 0.000299 moles of Ag+ is mixed with 0.023 moles of CN-
we say that essentially all of the 0.000299 moles of Ag+ is converted to [Ag(CN)2]^-1

the 0.000299 moles of [Ag(CN)2]^-1 has been diluted to a total volume of 345.0 ml:
(0.000299 moles of [Ag(CN)2]^-1) / (0.3450L) = 0.000103 Molar [Ag(CN)2]^-1

by the equation:
1 Ag+ & 2 CN^-1 --> [Ag+(CN-)2]^-1
twice as much CN- is consumed, when it reacts with the 0.000299 moles of Ag+:
(0.023 moles of CN-) - (2) (0.000299 moles lost) = 0.022402 mol CN- remains

which has also been diluted to 345.0 ml
(0.022402 mol CN- remains) / (0.345L) = 0.0649 Molar CN-

Kf = [Ag+(CN-)2] / [Ag+] [CN-]²

1 X 10²¹ = [0.000103 ] / [Ag+] [0.0649]²

[Ag+] = [0.000103] / (1 X 10²¹ ) [0.0649]²

[Ag+] = [0.000103] / (1 X 10²¹ ) (0.004212)

Ag = 2.44 x 10^-23 Molar