What volume of 0.5M H2SO4 will exactly neutralize 20cm3 of 0.1M NaOH Solution?

Analysis

To solve this problem, we need to use the principle of titration (volumetric analysis). We are given the molarity and volume of a base (NaOH) and the molarity of an acid ($H_2SO_4$). We need to find the volume of the acid required to neutralize the base.

Concept

The reaction between sulfuric acid ($H_2SO_4$) and sodium hydroxide ($NaOH$) is a neutralization reaction:

From the balanced equation, the mole ratio is:

• $n_a$ (moles of acid) = 1
• $n_b$ (moles of base) = 2

We use the titration formula:

Where:

• $C_a$ = Concentration of acid (0.5 M)
• $V_a$ = Volume of acid (?)
• $C_b$ = Concentration of base (0.1 M)
• $V_b$ = Volume of base (20 $cm^3$)

Solution

1. Plug the values into the formula: $$\frac{0.5 \times V_a}{0.1 \times 20} = \frac{1}{2}$$

2. Simplify the denominator: $$\frac{0.5 \times V_a}{2} = \frac{1}{2}$$

3. Cross-multiply or solve for $V_a$: $$2 \times (0.5 \times V_a) = 2 \times 1$$ $$1.0 \times V_a = 2$$ $$V_a = 2.0 \, cm^3$$

Conclusion

The volume of 0.5M $H_2SO_4$ required to neutralize the base is 2.0 $cm^3$.

Exam Tip

Always remember to balance your chemical equation first! Many students forget the mole ratio ($n_a/n_b$) and assume it is always 1:1. In this case, because $H_2SO_4$ is diprotic (has two $H^+$ ions), the ratio is 1:2. If you had used a 1:1 ratio, you would have gotten 4.0 $cm^3$, which is a common distractor in JAMB questions.