Solveeit Logo
Login

Question

Question: The concentration of hydrogen ion in a solution left after mixing 100 ml of 0.1 MgCl<sub>2</sub> and...

The concentration of hydrogen ion in a solution left after mixing 100 ml of 0.1 MgCl2 and 100 ml of 0.2 M NaOH. [KSP[Mg(OH)2]=12×10−11]\lbrack K_{SP}\left\lbrack Mg(OH)2 \right\rbrack = 12 \times 10^{- 11}\rbrack is

A

2.8×10−32.8 \times 10^{- 3}

B

2.8×10−22.8 \times 10^{- 2}

C

2.8×10−42.8 \times 10^{- 4}

D

2.8×10−52.8 \times 10^{- 5}

Answer

2.8×10−42.8 \times 10^{- 4}

Explanation

Solution

MgCl2+2NaOH→Mg(OH)2+2NaCl102000001020\begin{matrix} \mathbf{MgC}\mathbf{l}_{\mathbf{2}} & \mathbf{+} & \mathbf{2NaOH} & \mathbf{\rightarrow} & \mathbf{Mg}\left( \mathbf{OH} \right)_{\mathbf{2}} & \mathbf{+} & \mathbf{2NaCl} \\ \mathbf{10} & & \mathbf{20} & & \mathbf{0} & & \mathbf{0} \\ \mathbf{0} & & \mathbf{0} & & \mathbf{10} & & \mathbf{20} \end{matrix}

Thus 10 mili mole of Mg(OH)2 is formed

The ionic product

[Mg2+][OH−]2=[10200]×[20200]2=5×10−4\lbrack Mg^{2 +}\rbrack\left\lbrack OH^{-} \right\rbrack^{2} = \left\lbrack \frac{10}{200} \right\rbrack \times \left\lbrack \frac{20}{200} \right\rbrack^{2} = 5 \times 10^{- 4}which is more than Ksp of Mg(OH)2 .Now solubility (s) of Mg(OH)2 can be derived by

Ksp = 4s3

Mg(OH)2MG2+2OH−SS2 S\begin{array} { c c c } \mathrm { Mg } ( \mathrm { OH } ) _ { 2 } & \mathrm { MG } ^ { 2 + } & 2 \mathrm { OH } - \\ \mathrm { S } & \mathrm { S } & 2 \mathrm {~S} \end{array}

∴S=31.2×10−11=1.4×10−4∴[OH]=2s=2.8×10−4\therefore S = 3\sqrt{1.2 \times 10^{- 11}} = 1.4 \times 10^{- 4}\therefore\lbrack OH\rbrack = 2s = 2.8 \times 10^{- 4}