JEE MAIN - Chemistry (2023 - 13th April Morning Shift - No. 18)
$$20 \mathrm{~mL}$$ of calcium hydroxide was consumed when it was reacted with $$10 \mathrm{~mL}$$ of unknown solution of $$\mathrm{H}_{2} \mathrm{SO}_{4}$$. Also $$20 \mathrm{~mL}$$ standard solution of $$0.5 ~\mathrm{M} ~\mathrm{HCl}$$ containing 2 drops of phenolphthalein was titrated with calcium hydroxide, the mixture showed pink colour when burette displayed the value of $$35.5 \mathrm{~mL}$$ whereas the burette showed $$25.5 \mathrm{~mL}$$ initially. The concentration of $$\mathrm{H}_{2} \mathrm{SO}_{4}$$ is _____________ M. (Nearest integer)
Answer
1
Explanation
Reaction with HCl:
$$\mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O}$$
Volume of $\mathrm{Ca}(\mathrm{OH})_2=10 \mathrm{ml}$
Volume of $\mathrm{HCl}=20 \mathrm{ml}$
Concentration of $\mathrm{HCl}=0.5 \mathrm{M}$.
No. of milli moles of $\mathrm{HCl}=10$
No. of milli moles of $\mathrm{Ca}(\mathrm{OH})_2=5$.
i.e. $\mathrm{M}_{\mathrm{Ca}(\mathrm{OH})_2}=\frac{\text { no. of milli moles }}{\mathrm{V}(\mathrm{ml})}=\frac{5}{10}$ $=0.5 \mathrm{M}$.
Reaction with $\mathrm{H}_2 \mathrm{SO}_4$:
$$\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{H}_2 \mathrm{O} \text {. }$$
No. of milli moles of $\mathrm{Ca}(\mathrm{OH})_2=20 \times 0.5$ $=10$
i.e. no. of milli moles of $\mathrm{H}_2 \mathrm{SO}_4=10$
$$ \begin{aligned} \Rightarrow & \mathrm{M}_{\mathrm{H}_2 \mathrm{SO}_4}=\frac{\text { no, of milli moles }}{\mathrm{V}(\mathrm{ml})} \\\\ & =\frac{10}{10} \\\\ & =1 \mathrm{M} \end{aligned} $$
So, the concentration of $\mathrm{H}_2 \mathrm{SO}_4$ is 1 M.
$$\mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O}$$
Volume of $\mathrm{Ca}(\mathrm{OH})_2=10 \mathrm{ml}$
Volume of $\mathrm{HCl}=20 \mathrm{ml}$
Concentration of $\mathrm{HCl}=0.5 \mathrm{M}$.
No. of milli moles of $\mathrm{HCl}=10$
No. of milli moles of $\mathrm{Ca}(\mathrm{OH})_2=5$.
i.e. $\mathrm{M}_{\mathrm{Ca}(\mathrm{OH})_2}=\frac{\text { no. of milli moles }}{\mathrm{V}(\mathrm{ml})}=\frac{5}{10}$ $=0.5 \mathrm{M}$.
Reaction with $\mathrm{H}_2 \mathrm{SO}_4$:
$$\mathrm{Ca}(\mathrm{OH})_2+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CaSO}_4+2 \mathrm{H}_2 \mathrm{O} \text {. }$$
No. of milli moles of $\mathrm{Ca}(\mathrm{OH})_2=20 \times 0.5$ $=10$
i.e. no. of milli moles of $\mathrm{H}_2 \mathrm{SO}_4=10$
$$ \begin{aligned} \Rightarrow & \mathrm{M}_{\mathrm{H}_2 \mathrm{SO}_4}=\frac{\text { no, of milli moles }}{\mathrm{V}(\mathrm{ml})} \\\\ & =\frac{10}{10} \\\\ & =1 \mathrm{M} \end{aligned} $$
So, the concentration of $\mathrm{H}_2 \mathrm{SO}_4$ is 1 M.
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