Chapter 19: Q19.71P (page 875)

The solubility of zinc oxalate is $7.9 \times 10^{- 3} M at18°C.$ Calculate its $K_{sp}$ .

Short Answer

Expert verified

The $K_{sp}$ of zinc oxalate is $6.2 \times 10^{- 5} .$

Step by step solution

Concept Introduction

The $Q_{s p}$ -ion-product expression is obtained by multiplying the concentrations of ions generated by the dissolution of a chemical. When a solution is saturated, the $Q_{sp}$ value is referred to as the $K_{s p}$ value (solubility-product constant).

${MX}_{2} {> >M}^{2 +} {+ 2X}^{-}$

We do not include the solid and liquid states in the $K_{sp}$ equations

$K_{s p} = [M^{2 +}] {[X^{-}]}^{2}$ .

$(S)$ is equal to the concentration of one mol of the ion formed by dissolution of some compound.

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First, we can write the dissolution equation for ${ZnC}_{2} O_{4}$ :

$Z n C_{2} O_{4} (s) \Leftrightarrow Z n^{2 +} (a q) + C_{2} O_{4}^{2 -} (a q) S$

role="math" localid="1663309639326" $(Z n C_{2} O_{4}) = 7.9 \times 10^{- 3} M K_{s p} = [Z n^{2 +}] [C_{2} O_{4}^{2 -}] [Z n^{2 +}] = S [Z n^{2 +}] = 7.9 \times 10^{- 3} M$

$[C_{2} O_{4}^{2 -}] = S [C_{2} O_{4}^{2 -}] = 7.9 \times 10^{- 3} M K_{s p} = S \times S K_{s p} = 6.2 \times 10^{- 5} .$

Therefore, the $K_{s p}$ of zinc oxalate is $6.2 \times 10^{- 5} .$

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The solubility of zinc oxalate is $7.9 \times 10^{- 3} M at18°C.$ Calculate its $K_{sp}$ .

Short Answer

Step by step solution

Concept Introduction

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The solubility of zinc oxalate is7.9×10-3Mat18°C.Calculate itsKsp.

Short Answer

Step by step solution

Concept Introduction

information Provided

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Ionic and Molecular Compounds

Kinetics

Making Measurements

Inorganic Chemistry

Chemical Analysis

Study anywhere. Anytime. Across all devices.

The solubility of zinc oxalate is $7.9 \times 10^{- 3} M at18°C.$ Calculate its $K_{sp}$ .