Question

$0.05$ mole of $Ca{\left(OH\right)}_2$ can neutralise $H_{2}S{O}_4$. This $H_{2}S{O}_4$ can be neutralised by :

• A. $0.05$ mol of $NaOH$
• B. $0.10$ mol of $NaOH$
• C. $0.05$ mol of $POH{\left(OH\right)}_2$
• D. none of the above is correct

Answer 1

Answer: (B)

$0.10$ mol of $NaOH$

The balanced reaction is as follows:
$\underset{0.05mol}{Ca{\left(OH\right)}_2}+H_{2}S{O}_4⟶CaS{O}_4+2H_{2}O$
So according to balanced reaction,
$1molCa{\left(OH\right)}_2\equiv 1mol{H}_{2}S{O}_4$
$0.05molCa{\left(OH\right)}_2\equiv 0.05mol{H}_{2}S{O}_4$

$H_{2}S{O}_4+2NaOH\to N{a}_{2}S{O}_4+2H_{2}O$
1 2
0.05 0.10
So, $H_{2}S{O}_4$ can be neutralised by $0.10$ mol of $NaOH$.