PHYSICAL CHEMISTRY IN BRIEF

CHAP. 11: ELECTROCHEMISTRY [CONTENTS] 394

Solution

a) We have

Ag 2 CrO 4 (s) = 2 Ag++ CrO^24 −.

From the material balance it follows thatcAg+= 2candcCrO 24 −=c, wherecis the concentration

of the dissolved chromate. By substituting into (11.107) we obtain

9 × 10 −^12 = (2×c)^2 ×c/(cst)^3 =⇒ c= 1. 31 × 10 −^4 mol dm−^3.

b) We use the preceding result to calculate the ionic strength of the solution, see equations

(11.42)

I=

1

2

(2×c× 12 +c× 22 ) = 3×c= 3× 1. 37 × 10 −^4 = 4. 11 × 10 −^4.

We use relation (11.45) to calculate the mean activity coefficient

lnγ±=− 1. 176 × 1 × 2 ×

√

4. 11 × 10 −^4 =− 0 .0477 =⇒ γ±= 0. 953.

Now we substitute the mean activity coefficient into (11.107)

9 × 10 −^12 = (2×c)^2 ×c× 0. 9533 /(cst)^3 =⇒ c= 1. 37 × 10 −^4 mol dm−^3.

If we add to a solution of a sparingly soluble salt an electrolyte that shares a cation or an
anion with the salt, the solubility of the salt will decrease. If we add a small amount of an
electrolyte that does not share an ion with the salt, the solubility of the salt will increase.

Example

Calculate the solubility of silver chromate in a solution of potassium chromate of the concentration

cs= 0. 001 mol dm−^3. Use the data from the preceding Example and compare the results with its

solubility in pure water.