Calculate the volume of 0.5 mol dm\(^{-3}\) H\(_2\)SO\(_4\) that is neutralized by 25 cm\(^3\) of...

Calculate the volume of 0.5 mol dm\(^{-3}\) H\(_2\)SO\(_4\) that is neutralized by 25 cm\(^3\) of 0.1 mol dm\(^3\) NaOH?

Firstly, write a balanced equation for the chemical reaction: H\(_2\)SO\(_4\) + 2NaOH → Na\(_2\)SO\(_4\) + 2H\(_2\)O

C\(_A\) = 0.5 mol dm\(^{-3}\) V\(_A\) = ? C\(_B\) = 0.1 mol dm\(^{-3}\) V\(_B\) = 25 cm\(^3\)

N\(_A\) = 1 N\(_B\) = 2

\(\frac{{C_A}{V_A}}{{C_B}{V_B}}\) = \(\frac{N_A}{N_B}\)

V\(_A\) = \(\frac{{C_B}{V_B}{N_A}}{{C_A}{N_B}}\)

V\(_A\) = \(\frac{{0.1}\times{25}\times{1}}{{0.5}\times{2}}\)

V\(_A\) = 2.5 cm\(^3\)

There is an explanation video available below.

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