Question

A 125.0 −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 0.12 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains? Answer in M Express your answer using two significant figures. PS: the answer is not 2.9*10^{-20}

Answer 1

Ag+ + 2 CN-1 --> [Ag+(CN-)2]^-1

Number of moles of Ag+ in 125ml (0.125 L) of sample solution = 0.125L * 2.9×10−3 M = 0.0003625 moles of Ag+

Number of moles of CN- in 230 ml (0.230 L) sample solution = 0.230L * 0.12M in NaCN = 0.0276 moles of CN-

When 0.0003625 moles of Ag+ are mixed with 0.0276 moles of CN-, all of Ag+ is converted to [Ag(CN)2]^-1

The 0.0003625 moles of [Ag(CN)2]^-1 have been diluted to a total volume of 355 ml (0.355 L).

Therefore,

(0.0003625 moles of  [Ag(CN)2]^-1) / 0.355 L = 0.00102113 M  [Ag(CN)2]^-1

From the chemical equation, it is interpreted that twice CN- is consumed when it reacts with Ag+

(0.0276 moles of CN-) - (2) (0.0003625 moles lost) = 0.026925 mol CN- remains

which is diluted to 355.0 ml

(0.026925 mol CN-) / 0.355 L = 0.07584597 M CN-

Using equilibrium equation

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

1 X 10^21 = 0.00102113 / [Ag+] * [0.07584597]^2

[Ag+] = 0.00102113 / 1 X 10^21 * [0.07584597]^2

[Ag+] = 0.00102113 / 1 X 10^21 * 0.00575261

[Ag+] = 1.77 X 10^-22 M