What volume of 0.5 mol dm\(^{-3}\) H\(_2\)SO\(_4\) will exactly neutralize 20cm\(^3\) of 0.1 mol dm\(^{-3}\)...

What volume of 0.5 mol dm\(^{-3}\) H\(_2\)SO\(_4\) will exactly neutralize 20cm\(^3\) of 0.1 mol dm\(^{-3}\) NaOH solution?

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

From the balanced equation, N\(_A\) = 1 ; N\(_B\) = 2

C\(_A\) = 0.5 moldm\(^{-3}\) ; V\(_A\) = ? ; C\(_B\) = 0.1 moldm\(^{-3}\) ; V\(_B\) = 20 cm\(^3\)

Using the formula, \(\frac{{C_A}{V_A}}{{C_B}{V_B}}\) = \(\frac{N_A}{N_B}\)

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

= \(\frac{{0.1}\times{20}\times{1}}{{0.5}\times{2}}\)

V\(_A\) = 2.0 cm\(^3\)

There is an explanation video available below.

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