Question

50 gm of a sample of $Ca\left({OH}_2\right)$ is dissolved in 50 ml of 0.5 N HCI solution. The excess of HCI was titrated with $0.3NNaOH$. The volume of NaOH used was 20cc. Calculate % purity of $Ca\left({OH}_2\right)$ Fill your answer as sum of digits (excluding decimal places) till you get the single digit answer.

Answer 1

Volume of HCl $=\frac{50mL}{1000mL/L}=0.050L$
Number of moles of HCl $=0.050L\times 0.5N=0.025mol$

Volume of NaOH $=\frac{20cc}{1000cc/L}=0.020L$
Number of moles of NaOH $=0.020L\times 0.3N=0.006mol$

0.006 moles of NaOH will neutralise 0.006 moles of HCl.

Number of moles of HCl that have reacted with $Ca(OH{)}_2=0.025-0.006=0.019mol$

$Ca(OH{)}_2+2HCl\to CaC{l}_2+2H_{2}O$
Number of moles of $Ca(OH{)}_2=\frac{1}{2}\times$ number of moles of HCl.

Number of moles of $Ca(OH{)}_2=\frac{1}{2}\times 0.019=0.0095mol$

Molar mass of $Ca(OH{)}_2=74g/mol$
Mass of $Ca(OH{)}_2=0.0095mol\times 74g/mol=0.703g$

Mass of sample $=50g$

Percent purity of $Ca(OH{)}_2=\frac{0.703}{50}\times 100=1.4\%$
Answer as sum of digits (excluding decimal places) is 1.