Question

A 130.0 mL sample of a solution that is 2.7×10−3 M in AgNO₃ is mixed with a 230.0 mL sample of a solution that is 0.14 M in NaCN.

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

Answer 1

130.0 mL sample of a solution that is 2.7 x 10^-3 M in AgNO₃ is---( we will find the moles of Ag⁺) ---
(0.130 L )* (2.7x 10^-3 mol / L) = 0.000351 moles of Ag⁺

and 230.0 mL sample of a solution that is 0.14 M in NaCN is--( we will find the moles of CN^- )---
(0.230 L) * (0.14 mol /L ) = 0.0322 moles of CN-

When 0.000351 moles of Ag⁺ is mixed with 0.0322 moles of CN^- ,then we can assume that necessarily all of the 0.000351 moles of Ag⁺ has been converted to [Ag(CN)₂]^-1 .

Again, 0.000351 moles of [Ag(CN)₂]^-1 has been diluted to a total volume of (130.0 mL + 230.0 mL) = 360.0 mL
Now,

(0.000351 moles of [Ag(CN)₂]^-1) / (0.3600L) = 0.000975 M [Ag(CN)₂]-1

From the equation ,
1 Ag⁺ + 2 CN^-1 --> [Ag⁺(CN^-)₂]^-1
This shows that twice as much CN^- is consumed when it reacts with the 0.000351 moles of Ag⁺
(0.0322 moles of CN^-) - [2 x (0.000351 moles lost) ]

= (0.0322 - 0.000702) mol CN^- remains

=0.031498 mol CN^- remains.

And which has also been diluted to (130.0 + 230.0) mL =360.0 mL = 0.3600 L
(0.031498 mol CN^- remains) / (0.3600 L) = 0.0875 M CN^-

Finally,
K_f = [Ag(CN)2]^-1 / [Ag⁺] [CN^-]²

=> 1 X 10²¹ = [0.000975] / [Ag⁺] [0.0875]²

=> [Ag⁺] = [0.000975] / (1 X 10²¹ ) [0.0875]²

=> [Ag⁺] = [0.000975] / (1 X 10²¹ ) (0.007656)

=> [Ag⁺] = (0.000975 / 0.007656) x 10^-21

=> [Ag⁺] = 1.27 x 10^-1 x 10^-21

=> [Ag⁺] = 1.27 x 10^-22 M