Chapter 6: Q20P (page 142)

A solution contains $0.010 M {Ba}^{2 +}$ and $0.010 M {Ag}^{2}$ . Can 99.90% of either ion be precipitated by chromate ${CrO}_{4}^{2 -}$ without precipitating the other metal ion?

Short Answer

Expert verified

Neither ${Ba}^{2 +}$ nor ${Ag}^{+}$ can be precipitated without precipitating other metal ion.

Step by step solution

Chromate Precipitate.

${BaCrO}_{4 (s)} ⇌ {Ba}^{2 +} + {CrO}_{4}^{2 -} K_{sp} = 2.1 \times 10^{- 10} {Ag}_{2} {CrO}_{4 (s)} ⇌ 2 {Ag}^{+} + {CrO}_{4}^{2 -} K_{sp} = 1.2 \times 10^{- 12}$

Since, the stoichiometry of the two salts are not same, lower $K_{sp}$ value does not indicate faster precipitation.

Suppose 99.99% of ${BaCrO}_{4}$ is precipitated, then the final concentration of ${Ba}^{+ 2}$ ions will be $\frac{0.1}{100} \times 0.01 = 1.0 \times 10^{- 5} M$

Therefore, concentration of chromate ion present in solution :

$K_{sp} = [{Ba}^{2 +}] [{CrO}_{4}^{2 -}] 2.1 \times 10^{- 10} = 1.0 \times 10^{- 5} \times [{CrO}_{4}^{2 -}] [{CrO}_{4}^{2 -}] = \frac{2.1 \times 10^{- 10}}{1.0 \times 10^{- 5}} = 2.1 \times 10^{- 5}$

Reaction Quotient for Ag2CrO4 .

$Q = [{Ag}^{+}] [{CrO}_{4}^{2 -}] = {(0.010)}^{2} (2.1 \times 10^{- 5}) = 2.1 \times 10^{- 9}$

Since, $Q > K_{sp}$ for ${Ag}_{2} {CrO}_{4}, {Ag}^{+}$ should precipitate.

Therefore, it is not possible to separate both the salts by precipitation.

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A solution contains $0.010 M {Ba}^{2 +}$ and $0.010 M {Ag}^{2}$ . Can 99.90% of either ion be precipitated by chromate ${CrO}_{4}^{2 -}$ without precipitating the other metal ion?

Short Answer

Step by step solution

Chromate Precipitate.

Reaction Quotient for Ag2CrO4 .

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A solution contains 0.010MBa2+and 0.010MAg2. Can 99.90% of either ion be precipitated by chromate CrO42-without precipitating the other metal ion?

Short Answer

Step by step solution

Chromate Precipitate.

Reaction Quotient for Ag2CrO4 .

One App. One Place for Learning.

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A solution contains $0.010 M {Ba}^{2 +}$ and $0.010 M {Ag}^{2}$ . Can 99.90% of either ion be precipitated by chromate ${CrO}_{4}^{2 -}$ without precipitating the other metal ion?