Question

Step-I: 0.5 L of saturated $Mg(OH{)}_2$ (aq.) is in contact with $Mg(OH{)}_2$(s).
Step-II: 0.5 L of $H_{2}O$ is added to 0.5 L of solution in step (I) and the solution is vigorously stirred, undissolved $Mg(OH{)}_2$(s) remains.
Step-III: 100 mL of the clear solution in step (II) is removed and added to 0.5 L of 0.1 M $HCl$(aq.).
Step-IV: 25 mL of the clear solution in step (II) is removed and added to 225 mL of 0.06 M $MgC{l}_2$(aq.) $(K_{sp},Mg(OH{)}_2=3.2\times {10}^{-11})$.
Which of the following are correct choices about above steps?

• A. $\text{Concentration of}M{g}^{2+}\text{(aq.) in step (II) is}2\times {10}^{-4}M$
• B. $\text{Concentration of}M{g}^{2+}\text{(aq.) in step (III) is}3.33\times {10}^{-5}M$
• C. $\text{Concentration of}M{g}^{2+}\text{(aq.) in step (IV) is nearly equal to}0.054M$
• D. $\text{Precipitation of}Mg\left(OH{)}_2\right(s)\text{will occurs in step (III)}$

Answer 1

Answer: (A), (B), (C)

$\text{Concentration of}M{g}^{2+}\text{(aq.) in step (IV) is nearly equal to}0.054M$

Step-I: $\left[M{g}^{2+}\right]=\sqrt[3]{3\frac{3.2\times {10}^{-11}}{4}}=2\times {10}^{-4}M$
Step-II: $\left[M{g}^{2+}\right]=2\times {10}^{-4}M$ since it will again gives saturated solution.
Step-III: $\left[M{g}^{2+}\right]=2\times {10}^{-4}\times \frac{100}{(100+500)}=3.33\times {10}^{-5}M$
Step-IV: $Mg(OH{)}_2\text{(25 mL of}2\times {10}^{-4}\text{M or 0.005 m mole})$ is mixed with $MgC{l}_2$ (225 mL of 0.06 M or 13.5 m mole)
so $[M{g}^{2+}{]}_{\text{mainly from}MgC{l}_2}\simeq \frac{13.5}{250}M=0.054M$