- #1

- 3,812

- 92

## Homework Statement

How much AgBr could dissolve in 1.0 L of 0.40 M NH

_{3}? Assume that Ag(NH

_{3})

_{2}

^{+}is the only complex formed. [K

_{f}=1*10

^{8}; K

_{sp}=5*10

^{-13}]

## Homework Equations

## The Attempt at a Solution

Let Ag

^{+}and Br

^{-}formed from AgBr be x. Since Ag

^{+}form a complex with NH

_{3}, let the amount of Ag+ remaining be x-y, hence amount of NH

_{3}remaining is 0.4-2y and amount of complex formed is y.

Therefore

[tex]K_{sp}=(x-y)x[/tex]

[tex]K_f=\frac{y}{(x-y)(0.4-2y)^2}[/tex]

I hope my equations are correct.

Now the trouble is solving the equations. I can easily plug them in a calculator and get the answer but I want to know if there is a way to do this problem without using one.

Since K

_{f}is so large, I think that ammonia will be almost completely used and hence, y=0.2 M but I don't see if this helps or if my assumption is correct.

Any help is appreciated. Thanks!

Last edited: