When $1 L$ of 0.1M sulphuric acid solution is allowed to react with $1 L$ of $0.1 M$ sodium hydroxide solution, the amount of sodium sulphate (anhydrous) that can be obtained from the solution formed and the concentration of $H^{+}$ in the solution respectively are:

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$3.55 g, 0.025 M$

$7.10 g, 0.05 M$

$3.55 g, 0.1 M$

$7.10 g, 0.025 M$

answer is D.

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Detailed Solution

In the reaction,

$\begin{array}{l} 0.1 M, 1 {LH}_{2} {SO}_{4} + 1 L, 0.1 MNaOH ⟶ {Na}_{2} {SO}_{4} \\ 2 NaOH + H_{2} {SO}_{4} ⟶ {Na}_{2} {SO}_{4} + 2 H_{2} O \\ 0.1 \times 1 0.1 \times 1 0.05 M \end{array}$

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