QFE Is it possible to precipitate 99... [FREE SOLUTION]

Chapter 0: QFE (page 1)

Is it possible to precipitate $99.0 %$ of role="math" localid="1663573455203" $0.010 {MCe}^{3 +}$ by adding oxalate $(C_{2} {O^{2}}_{4})$ without precipitating $0.010 {MCe}^{2 +}$ ?
role="math" localid="1663574065069" ${CaC}_{2} O_{4} K_{sp} = 1.3 脳 10^{_{- 8}} {Ce}_{2} (C 2 O 4) 3 K_{sp} = 5.9 脳 10^{- 30}$

Short Answer

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Step by step solution

Step 1:

The possibility to precipitate $99 %$ of $0.010 {MCe}^{3 +}$ by adding oxalate without precipitating $0.010 {MCe}^{2 +}$ must be predicted.

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Given

${CaC}_{2} O_{4} K_{sp} = 1.3 脳 10^{_{- 8}} {Ce}_{2} (C 2 O 4) 3 K_{sp} = 5.9 脳 10^{- 30}$

There is a need to reduce concentration of ${Ce}^{3 +}$ to $1.0 % of$ $0.010 M (= 0.00010 M)$ . The concentration of oxalate in equilibrium WITH $0.00010 {MCe}^{3 +}$ is calculated as follows,

${[{Ce}^{3 +}]}^{2} [C_{2} O_{4}^{2}]^{3} = Ksp " = 5.9 脳 10^{- 30}$

${(0.00010)}^{2} [C_{2} O_{4}^{2}]^{3} = 5.9 脳 10^{- 30}$

[C_{2} O_{4}^{2 -}] = (\frac{5.9 脳 脳 10^{- 30}}{{(0.0001)}^{2}}) = 8.4 脳 10^{- 8} M

If $8.4 脳 10^{- 8} {MC}_{4}^{2 -}$ will precipitate $0.010 {MCa}^{2 +}$ , calculate $Q$ for ${CaC}_{2} O_{4}$ :

$[{Ca}^{2 +}] = [C_{2} O_{4}^{2 -}] = (0.010) (8.4 脳 10^{- 8}) = 8.4 脳 10^{- 8}$

Because $Q < K_{sp}$ for ${CaC}_{2} O_{4} = 1.3 脳 10^{- 8} {Ca}^{2 +}$ , so will not precipitate.

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Step by step solution

Step 1:

check for precipitate

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91影视

Is it possible to precipitate 99.0%of role="math" localid="1663573455203" 0.010MCe3+by adding oxalate (C2O24)without precipitating 0.010MCe2+?role="math" localid="1663574065069" CaC2O4Ksp=1.3脳10-8Ce2(C2O4)3Ksp=5.9脳10-30

Short Answer

Step by step solution

Step 1:

check for precipitate

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Kinetics

Chemical Analysis

Ionic and Molecular Compounds

The Earths Atmosphere

Organic Chemistry

Chemistry Branches

Study anywhere. Anytime. Across all devices.