Check out some similar questions!

hslove142331 · Apr 6, 2010, 02:31 PM

At 25°C, the value of Ksp for AgCN is 2.2 x 10-16 and that for Zn(CN)2 is 3.0 x 10-16.

In units of grams per 100 mL of solution, what is the solubility of AgCN?

In units of grams per 100 mL of solution, what is the solubility of Zn(CN)2?

I tried so many times to do this problem, but I think I have some problem with how to do this problems.

Please help me that how can I do like this problems?

Thanks

Unknown008 · Apr 7, 2010, 03:27 AM

First, write down the Ksp expression.

I'll do that of AgCN and you'll try that for Zn(CN)2.

$AgCN \rightleftharpoons Ag^+ + CN^-$

$K_{sp} = [Ag^+][CN^-]$

That is:

$2.2 \times 10^{-16} = [Ag^+][CN^-]$

Look at the equation now. Let's take one mole/dm^3 of initial AgCN (I always use one mole for the ease of calculations, and this was, we also have the AgCN largely in excess)

Initial AgCN: 1 M, Ag+ : 0 M, CN- : 0 M
At equilibrium, AgCN: 1-x M, Ag+ : x M, CN- : x M

If you had x moles from the initial AgCN dissociating, then you have 1-x moles of AgCN left, and you get x moles of each Ag+ and CN-.

Plug that in the equation.

$2.2 \times 10^{-16} = [Ag^+][CN^-]$

$2.2 \times 10^{-16} = [x][x]$

$2.2 \times 10^{-16} = x^2$

$x = \sqrt{2.2 \times 10^{-16}} = 1.48 \times 10^{-8}$

This means that 1.48x10^-8 M dissociated completely.

Now, you have the concentration that dissolves (dissociates), convert it into grams per Litre.

1.48 x 10^-8 = (107.9+12+14) * 1.48 x 10^-8 = 1.99 x 10^-6 g/L

For 1000 mL, it is 1.99 x 10^-6 g
For 100 mL, it will be 1.99 x 10^-7 g

Hence, solubility = 1.99 x 10^-7 g/ 100 mL