Question

$50$ g of a sample of $Ca{\left(OH\right)}_2$ is dissoved in $50$ mL of $0.5N$ $HCl$ solution. The excess of $HCl$ was titrated with $0.3NNaOH$. The volume of $NaOH$ used was $20$ cc. The percentage purity of $Ca{\left(OH\right)}_2$ is:

• A. $1.406\%$
• B. $5.406\%$
• C. $1.506\%$
• D. $1.45\%$

Answer 1

Answer: (A)

$1.406\%$

Meq. of $Ca{\left(OH\right)}_2=$ Meq. of $HCl$

$\left(\frac{w}{\frac{74}{2}}\right)\times 1000=(50\times 0.5-0.3\times 20)$ $⟹w=0.703$

Percentage of $Ca{\left(OH\right)}_2=\frac{w}{50}\times 100=\frac{0.703}{50}\times 100=1.406\%$