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

Explanation

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\(^3\)

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