Complex ions?

Question

what is the conc. of Ag+ in 0.0560 M AgNO3 and 1.1823 M CN-?

Ag+ + 2CN- <-- --> [Ag(CN)2]- Kf=5.3x10^18

0____1.1823M____0.0560M

+x_____+2x_________-x

x___1.1823+2x____0.0560-x

Did i set up this table right??

Kf = 5.3x10^18 = x*(1.1823+2x) / (.0560-x)

Am i solving this right? How do i solve the above Kf equation??

Answer 1

You&#039;re close except for a few details.

Since Kf is so large, we assume that all the Ag+ from AgNO3 reacts with CN- to form Ag(CN)2-. In one liter of solution, 0.0560 moles of Ag+ will react with 2 x 0.0560 = 0.112 moles of CN- to form 0.0560 moles of Ag(CN)2-. The amount of CN- remaining is 1.1823 moles - 0.112 moles = 1.0703 moles CN-.

____Ag(CN)2-   &lt;++&gt;   Ag+  +  2CN-

____0.0560 M________0_____1.0703 M

_____-x_____________+x_____+2x____

____0.0560-x________+x_____1.0703+2x

Kd = 1/Kf = 1.9 x 10^-20 = [Ag+][CN-]^2 / [Ag(CN)2-] = (x)(1.0703 + 2x) / (0.0560 - x)

Because Kd is SO small, it is negligible compared to 0.0560 and 1.0703, so we can ignore the +2x and -x in the above equation and greatly simplify things.

1.9 x 10^-20 = 1.0703x / 0.0560 = 19.11x

x = 9.9 x 10^-22 M = [Ag+]