What volume of 0.5mldm3 H2SO4 will exactly neutralize 20cm3 of 0.1moldm3 NaOH Solution?

Analysis

To solve this problem, we need to use the principle of titration (neutralization). We are given the concentration and volume of a base (NaOH) and the concentration 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:

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 = $0.5 \, \text{mol/dm}^3$
• $V_a$ = Volume of acid = ?
• $C_b$ = Concentration of base = $0.1 \, \text{mol/dm}^3$
• $V_b$ = Volume of base = $20 \, \text{cm}^3$

Solution

1. Substitute 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 to solve for $V_a$: $$2 \times (0.5 \times V_a) = 2 \times 1$$ $$1 \times V_a = 2$$ $$V_a = 2.0 \, \text{cm}^3$$

Conclusion

The volume of $0.5 \, \text{mol/dm}^3$ $H_2SO_4$ required is $2.0 \, \text{cm}^3$.

Exam Tip

Always remember to balance your chemical equation first! Many students forget the mole ratio ($n_a/n_b$) and simply use $C_aV_a = C_bV_b$. If you ignore the stoichiometry of the reaction, you will get the wrong answer. In this case, the $H_2SO_4$ is diprotic (has two $H^+$ ions), which is why the ratio is 1:2.