1f:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 20:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1e:T443,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"20 mL of calcium hydroxide was consumed when it was reacted ","item":"https://www.divinejee.com/question/20-ml-of-calcium-hydroxide-was-consumed-when-it-was-reacted-with-10-ml-of-unknown-solution-of-h2-so4"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question 20 mL of calcium hydroxide was consumed when it was reacted on Topic","text":"

$20 \\mathrm{~mL}$ of calcium hydroxide was consumed when it was reacted with $10 \\mathrm{~mL}$ of unknown solution of $\\mathrm{H}_{2} \\mathrm{SO}_{...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/20-ml-of-calcium-hydroxide-was-consumed-when-it-was-reacted-with-10-ml-of-unknown-solution-of-h2-so4"}}}]}21:T4a0,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.