# Solubility Product

## Solubility Product (K_{sp})

Solubility product is defined as- in the saturated solution, the product of concentrations of the ions raised to a power equal to the number of times, the ions occur in the equation representing the dissociation of the electrolyte at a given temperature.Consider in general, the electrolyte of the type A

_{x}B

_{y}which is dissociated as-

A

_{x}B

_{y}⇌ xA

^{+y}+ yB

^{−x}

Applying law of mass action-

K = [A

^{+y}]

^{x}[B

^{−x}]

^{y}/[A

_{x}B

_{y}]

or, K [A

_{x}B

_{y}] = [A

^{+y}]

^{x}[B

^{−x}]

^{y}

for saturated solution-

[A

_{x}B

_{y}] = constant

so, K

_{sp}= [A

^{+y}]

^{x}[B

^{−x}]

^{y}

### Q. The solubility of barium sulphate at 298 K is 1.05 x 10^{-5} mol dm^{-3}. Calculate the solubility product.

BaSO_{4}⇌ Ba

^{+2}+ SO

_{4}

^{-2}

Given-

[Ba

^{+2}] = 1.05 x 10

^{-5}mol dm

^{-3}

[SO

_{4}

^{-2}] = 1.05 x 10

^{-5}mol dm

^{-3}

So, K

_{sp}= [Ba

^{+2}] [SO

_{4}

^{-2}]

or, K

_{sp}= 1.05 x 10

^{-5}X .05 x 10

^{-5}

or, K

_{sp}= 1.10 x 10

^{-10}mol

^{2}dm

^{-6}

### Q. If the solubility product of magnesium hydroxide is 2.00 X 10^{-11} mol^{3} dm^{-9} at 298 K, calculate its solubility in mol dm^{-3} at that temperature.

Answer: 1.71 X 10^{-4}mol dm

^{-3}

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