- #1

tobywashere

- 28

- 0

## Homework Statement

What is the minimum amount of AgNO

_{3}, in grams, that has to be added to 2.50L of 0.020M K

_{2}CrO

_{4}to produce a precipitate?

## Homework Equations

K

_{sp}Ag

_{2}CrO

_{4}= 9.0 x 10

^{-12}

[Ag

^{+}]

^{2}[CrO

_{4}

^{-2}] = 9.0x10

^{-12}

## The Attempt at a Solution

2AgNO

_{3}(aq) + K

_{2}CrO

_{4}(aq) --> Ag

_{2}CrO

_{4}(s) + 2KNO

_{3}(aq)

Ag

_{2}CrO

_{4}(s) is the precipitate

The net ionic equation is:

Ag

_{2}CrO

_{4}(s) --> 2Ag

^{+}(aq) + CrO

_{4}

^{-2}(aq)

Let y represent the concentration of CrO

_{4}

^{-2}at equilibrium. The concentration of Ag

^{+}will be 2y due to the 2:1 stoichiometry. Plugging into the solubility equilibrium expression:

[Ag

^{+}]

^{2}[CrO

_{4}

^{-2}] = 9.0x10

^{-12}

(2y)

^{2}(y) = 9.0x10

^{-12}

Therefore, y = 1.31 x 10

^{-4}M

So the equilibrium concentration of CrO

_{4}

^{-2}is 1.31 x 10

^{-4}M, and the equilibrium concentration of Ag

^{+}is double that, or 2y, or 2.62x10

^{-4}M

The initial concentration of CrO

_{4}

^{-2}is 0.020M (from the 0.020M K

_{2}CrO

_{4}). The decrease in concentration of CrO

_{4}

^{-2}to form the precipitate is:

0.020M - 1.31 x 10

^{-4}M = 0.01987M

The initial concentration of Ag

^{+}(what we're trying to find) decreases by twice this amount, 0.03974M, to reach an equilibrium concentration of 2.62x10

^{-4}M (the 2y found earlier).

Initial concentration of Ag

^{+}- 0.03974M = 2.62x10

^{-4}M.

Initial concentration = 2.62x10

^{-4}M + 0.0397M = 0.040M

Find mass:

0.040M x 2.5L x 170g / mol = 17g

Therefore, 17 g of AgNO

_{3}needs to be added to form a precipitate.

Did I do this correctly? I'm really not sure.<br>