A chemist measures the solubility of lead(ii) bromide in water to be 2. 96 g/l. calculate the ksp value for the solid. the ksp v

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zimovet [89]

2 years ago

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A chemist measures the solubility of lead(ii) bromide in water to be 2. 96 g/l. calculate the ksp value for the solid. the ksp v

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Chemistry

1 answer:

Julli [10] · Answer 1 · 2023-06-25T08:44:51+03:00

The solubility product, ksp of PbBr₂ is 2.102 × 10⁻⁶

<h3>What is solubility?</h3>

The solubility of a solute is defined as the maximum amount of that solute that can be dissolved in a known quantity of solvent at a given temperature.

<h3>What is a solubility product? </h3>

Some salts are sparingly soluble in a solvent. For them, we calculate the solubility product.

It is an equilibrium constant that defines the relationship between a solid and its respective ions in an aqueous solution in equilibrium.

The greater the solubility product, the greater the solubility and vice versa.

Here, the solubility of PbBr₂ = 2.96 g/l

Molar solubility of PbBr₂ = $\frac{2.96}{molar mass of PbBr_2}$ = 2.96/367 = 8.07 × 10⁻³

At equilibrium,

$PbBr_2\rightarrow Pb^{2+} + 2Br^-$

1 mole of PbBr2 dissociates into 1 mole of Pb²⁺ ions and 2 moles of Br⁻

Let the molar concentration of Pb²⁺ be x, then the molar concentration of Br⁻ is 2x

Ksp = x.(2x)²

= 4x³

Substitute, x = 8.07 × 10⁻³

Ksp = 4 (8.07 × 10⁻³)³

= 2.102 × 10⁻⁶

Thus, The ksp of PbBr₂ is 2.102 × 10⁻⁶

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