Question

A mixture of $NaOH$ and $Mg{\left(OH\right)}_2$ weights 2.325 g. It requires 3 g of $H_2{SO}_4$ for its neutralization. What is the percentage composition of mixture?

Answer 1

A mixture of NaOH and $mg(OH{)}_2$ weights $2.325g$

Let $W_{NaOH}=ag$

$W_{mg(OH{)}_2}=bg$

$\therefore a+b=\mathrm{2.325...}\left(i\right)$

For neutralization,

$Meq\left(NaOH\right)+Meq\left(Mg\right(OH{)}_2)=meq(H_{2}S{O}_4)$

$\therefore \frac{a\times 1000}{40}+\frac{b\times 1000}{58.3/2}=\frac{3\times 1000}{98/2}$

On solving,

$0.025a+0.034b=0.061...\left(2\right)$

On Solving, equation (i) and (ii) we get,

$a=w_{NaOH}=2g$

$b=W_{mg(OH{)}_2}=0.32g$

$\%NaOH=\frac{2}{2.325}\times 100=86\%$

$\%mg\left(O{H}_2\right)=\frac{0.32}{2.325}\times 100=13.76\%$