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

1. Analyze the question

We are asked to find the volume of hydrochloric acid (HCl) needed to neutralize a specific volume and concentration of sodium hydroxide (NaOH). This is a classic acid-base titration problem.

2. Explain the concept

To solve this, we use the titration formula:

Where:

• $C_a$ = Concentration of the acid (HCl)
• $V_a$ = Volume of the acid
• $n_a$ = Number of moles of acid in the balanced equation
• $C_b$ = Concentration of the base (NaOH)
• $V_b$ = Volume of the base
• $n_b$ = Number of moles of base in the balanced equation

The balanced chemical equation is:

From the equation, $n_a = 1$ and $n_b = 1$.

3. Show the solution

Given data:

• $C_a = 0.12 \text{ mol dm}^{-3}$
• $V_a = ?$
• $C_b = 0.20 \text{ mol dm}^{-3}$
• $V_b = 20 \text{ cm}^3$

Step-by-step calculation: Substitute the values into the formula:

Simplify the right side:

Solve for $V_a$:

4. Identify the answer

Rounding to two decimal places, we get 33.33 cm³. This matches option C.

5. Exam Tip

In titration questions, always check the stoichiometry (the coefficients) of the balanced equation first. If the acid were $H_2SO_4$ instead of $HCl$, the $n_a$ would be 1, but if you were dealing with a diprotic acid, you must ensure your $n_a$ and $n_b$ values are correct, or your calculation will be off by a factor of 2!