If the ionic product of M(OH)2 is 5 × 10-10, then the molar solubility of M(OH)2 in 0.1 M NaOH is

Question

If the ionic product of M(OH)2 is 5 × 10-10, then the molar solubility of M(OH)2 in 0.1 M NaOH is (A) 5 × 10-12 M (B) 5 × 10-8 M (C) 5 × 10-10 M (D) 5 × 10-9 M

Tardigrade Admin · Accepted Answer

Ionic product of M( OH 2=5 × 10-10. or [M2+][ OH -]2=5 × 10-10 Given [ OH -]=10-1 mol / L So, [M2+] =(5 × 10-10/[10-1]2) =(5 × 10-10/10-2)=5 × 10-8 M

Q. If the ionic product of $M (O H)_{2}$ is $5 \times 1 0^{- 10}$ , then the molar solubility of $M (O H)_{2}$ in $0.1 M N a O H$ is

$5 \times 1 0^{- 12}$ M

$5 \times 1 0^{- 8}$ M

$5 \times 1 0^{- 10}$ M

$5 \times 1 0^{- 9}$ M

$5 \times 1 0^{- 16}$ M

Ionic product of $M (O H_{2} = 5 \times 1 0^{- 10}$

or $[M^{2 +}] [O H^{-}]^{2} = 5 \times 1 0^{- 10}$

Given $[O H^{-}] = 1 0^{- 1} m o l / L$

So, $[M^{2 +}] = \frac{5 \times 1 0 ^{- 10}}{[ 1 0 ^{- 1} ] ^{2}}$

$= \frac{5 \times 1 0 ^{- 10}}{1 0 ^{- 2}} = 5 \times 1 0^{- 8} M$

Q. If the ionic product of M(OH)2​ is 5×10−10, then the molar solubility of M(OH)2​ in 0.1MNaOH is

5×10−12 M

5×10−8 M

5×10−10 M

5×10−9 M

5×10−16 M

Ionic product of M(OH2​=5×10−10 or [M2+][OH−]2=5×10−10 Given [OH−]=10−1mol/L So, [M2+]=[10−1]25×10−10​ =10−25×10−10​=5×10−8M