If 25 𝑚𝐿 \(𝐻_{2}𝑆𝑂_{4}\) solution reacts completely with \(1.06\) 𝑔 of pure \(𝑁𝑎_{2}𝐶𝑂_{3}\) , what is the normality of this acid solution?

For neutralisation reaction, \(𝐻_{2}𝑆𝑂_{4} + 𝑁𝑎_{2}𝐶𝑂_{3} → 𝑁𝑎_{2}𝑆𝑂_{4} + 𝐶𝑂_{2} + 𝐻_{2}𝑂\)

Equivalents of \(𝐻_{2}𝑆𝑂_{4}\) = Equivalents of \(𝑁𝑎_{2}𝐶𝑂_{3}\)

\(𝑁𝑜𝑟𝑚𝑎𝑙𝑖𝑡𝑦 × 𝑉𝑜𝑙𝑢𝑚e=\dfrac{Given mass}{Molar mass}\times n-factor\)

\(N\times\dfrac{25}{1000}=\dfrac{1.06}{106}\times 2\)

\(𝑁 = 0.8 ~N\)

Thus, the normality of the acid solution is \(0.8~N\)

So, option (D) is the correct answer.

Additional information

For \(𝑁𝑎_{2}𝐶𝑂_{3}\) , n factor = Total charge on cation or anion.

\(𝑁𝑎_{2}𝐶𝑂_{3} \rightleftharpoons 2𝑁𝑎~^{+} + 𝐶𝑂_{3}^{2-}\)

Thus 𝑛 − 𝑓𝑎𝑐𝑡𝑜𝑟 = 2