Question

$10$g $CaC{O}_3$ were dissolved in $250$ml of $1$M HCl. What volume of $2$M KOH would be required to neutralize excess HCl.

Answer 1

$CaC{O}_3+2HCl⟶H_{2}C{O}_3+CaC{l}_{2}10gm250ml$

In general $1$ mole of $CaC{O}_3$ reacts with $2$moles of HCl.

Number of moles of $\frac{10}{M_{CaC{O}_3}}=\frac{10}{100}=0.1$ mole.

Number of moles of HCl $=Molarity\times Volume=\frac{250}{1000}\times 1=0.25$ mole.

$0.1$ mole of $CaC{O}_3$ will react with $0.2$ moles of HCl(law of constant proportion).

So number of moles of HCl left $=0.25-0.2=0.05$ mole.

Number of equivalents of KOH =Number of equivalents of HCl

$M_1{V}_1=M_2{V}_2{2V}_1=0.05V_1=0.025l\_\_\_\_\_\_\_\_\_V=25ml$