Question

A 110.0 −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 0.11 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

110.0 mL sample of a solution that is 2.9×10⁻³ M in AgNO₃ is
(0.110 L ) (2.9×10⁻³ mol / L) = 0.000319 moles of Ag+

230.0 mL sample of a solution that is 0.11M in NaCN is
(0.230 L) (0.11 M in NaCN) = 0.0253 moles of CN-
when 0.000319 moles of Ag+ is mixed with 0.0253 moles of CN-
we say that essentially all of the 0.000319 moles of Ag+ is converted to [Ag(CN)2]^-1

then 0.000319moles of [Ag(CN)2]^-1 has been diluted to a total volume of 340.0 ml:
(0.000319 moles of [Ag(CN)2]^-1) / (0.340L) = 0.000938 Molar [Ag(CN)2]^-
(& you got that as 9.38e-4)
by the equation:
1 Ag+ & 2 CN-1 --> [Ag+(CN-)2]^-1
twice as much CN- is consumed, when it reacts with the 0.000319 moles of Ag+:
(0.0253 moles of CN-) - (2) (0.000319 moles lost) = 0.024662 mol CN- remains

which has also been diluted to 340.0 ml
(0.024662 mol CN- remains) / (0.340L) = 0.0725 Molar CN-

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

1 X 10^21 = [0.000938 ] / [Ag+] [0.0725]^2

[Ag+] = [0.000938] / (1 X 10^21 ) [0.0725]^2

[Ag+] = [0.000938] / (1 X 10^21 ) (0.00525)

[Ag+] = 1.78 X 10^-22 Molar