The volume in cm\(^3\) of a 0.12 moldm\(^{-3}\) HCl required to completely neutralize a 20cm\(^3\)...

The volume in cm\(^3\) of a 0.12 moldm\(^{-3}\) HCl required to completely neutralize a 20cm\(^3\) of 0.20 moldm\(^{-3}\) of NaOH is

First, write a balanced equation and ensure the stoichiometry is in order:

HCl + NaOH → NaCl + H\(_2\)O

Following the reaction above, the mole ratio is 1 : 1

Given:

Concentration of Acid, C\(_A\) = 0.12moldm\(^{-3}\), Volume of Acid, V\(_A\) = ?

Concentration of Base, C\(_B\) = 0.20 moldm\(^{-3}\) Volume of Base, V\(_B\) = 20cm\(^3\)

Using the relation, C\(_A\)V\(_A\) = C\(_B\)V\(_B\)

V\(_A\) = \(\frac{C_BV_B}{C_A}\)

= \(\frac{0.20 \times 20}{0.12}\)

= 33.33cm\(^3\)

There is an explanation video available below.

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