Question

How many millilitres of $0.1N{H}_{2}S{O}_4$ solution will be required for complete reaction with a solution containing $0.125g$ of pure $N{a}_{2}C{O}_3$?

• A. $23.6ml$
• B. $25.6ml$
• C. $26.3ml$
• D. $32.6ml$

Answer 1

Answer: (A)

$23.6ml$

$23.6mL$ of $0.1N$ $H_{2}S{O}_4$ solution will be required for complete reaction with a solution containing $0.125g$ of pure $N{a}_{2}C{O}_3$

For complete reaction no. of equivalents must be equal.

No. of moles of $N{a}_{2}C{O}_3=$ mass/molar mass $=0.125/106=0.001179$ mol

No. of equivalents of $N{a}_{2}C{O}_3=$no. of moles $x$ valence factor $=0.001179\times 2=$ $0.002358$

This is no. of equivalents $H_{2}S{O}_4$ must be present for complete reaction.

For $H_{2}S{O}_4$ no. of equivalents = $Normality\times volume$ (in litres)

Therefore volume in $L$ of $H_{2}S{O}_4$ $=0.002358/0.1=0.02358$ Litre

$=0.02358\times 1000mil=23.58$ ml

So, answer is $23.58$ ml be present for complete reaction.