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QJEEM23Chac86: JEE Main 2023 Chemistry - Class 11 JEEM23Chac86 Solution: of calcium hydroxide was consumed when it was reacted with of…

20 mL20 \mathrm{~mL} of calcium hydroxide was consumed when it was reacted with 10 mL10 \mathrm{~mL} of unknown solution of H2SO4\mathrm{H}_{2} \mathrm{SO}_{4} . Also 20 mL20 \mathrm{~mL} standard solution of 0.5 M HCl0.5 ~\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 mL35.5 \mathrm{~mL} whereas the burette showed 25.5 mL25.5 \mathrm{~mL} initially. The concentration of H2SO4\mathrm{H}_{2} \mathrm{SO}_{4} is _____________ M. (Nearest integer)

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Reaction with HCl:

Ca(OH)2+2HCl→CaCl2+2H2O\mathrm{Ca}(\mathrm{OH})_2+2 \mathrm{HCl} \rightarrow \mathrm{CaCl}_2+2 \mathrm{H}_2 \mathrm{O}

Volume of Ca(OH)2=10ml\mathrm{Ca}(\mathrm{OH})_2=10 \mathrm{ml}

Volume of HCl=20ml\mathrm{HCl}=20 \mathrm{ml}

Concentration of HCl=0.5M\mathrm{HCl}=0.5 \mathrm{M} .

No. of milli moles of HCl=10\mathrm{HCl}=10

No. of milli moles of Ca(OH)2=5\mathrm{Ca}(\mathrm{OH})_2=5 .

i.e. MCa(OH)2= no. of milli moles V(ml)=510\mathrm{M}_{\mathrm{Ca}(\mathrm{OH})_2}=\frac{\text { no. of milli moles }}{\mathrm{V}(\mathrm{ml})}=\frac{5}{10} =0.5M=0.5 \mathrm{M} .

Reaction with H2SO4\mathrm{H}_2 \mathrm{SO}_4 :

Ca(OH)2+H2SO4→CaSO4+2H2O. \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 Ca(OH)2=20×0.5\mathrm{Ca}(\mathrm{OH})_2=20 \times 0.5 =10=10

i.e. no. of milli moles of H2SO4=10\mathrm{H}_2 \mathrm{SO}_4=10

⇒amp;MH2SO4= no, of milli moles V(ml)amp;=1010amp;=1M\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 H2SO4\mathrm{H}_2 \mathrm{SO}_4 is 1 M.

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