First slide

Solubility product(KSP)

Question

$M {(OH)}_{X} has K_{SP} 4 \times 10^{- 12}$ and solubility $10^{- 4} M .$ The value of x is

Moderate

A

1
B

2
C

3
D

4

Solution

$M {(OH)}_{x} ⇌ M^{+ x} + {xOH}^{-} s xs k_{sp} = s {(xs)}^{x} 4 \times 10^{- 12} = 10^{- 4} {(x 10^{- 4})}^{x} x^{x} \times 10^{- 4 x} = 4 \times 10^{- 8}; 2^{2} \times 10^{- 4 \times 2} = 4 \times 10^{- 8}$
By putting x=2 the equation is satisfied. Hence x = 2

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