Calculate the pH of 0.05 moldm$^{-3}$ H$_2$SO$_4$

Analysis

To calculate the pH of a solution, we need to determine the concentration of hydrogen ions $[H^+]$ in the solution. Sulfuric acid ($H_2SO_4$) is a strong diprotic acid, meaning it dissociates completely in water to release two hydrogen ions per molecule.

Concept

The formula for pH is:

Since $H_2SO_4$ is a diprotic acid, the dissociation equation is:

This means the concentration of hydrogen ions $[H^+]$ is twice the concentration of the acid:

Step-by-Step Solution

1. Identify the concentration of the acid: $[\text{Acid}] = 0.05 \text{ mol dm}^{-3}$

2. Calculate the concentration of hydrogen ions $[H^+]$: Since each mole of $H_2SO_4$ produces 2 moles of $H^+$, $[H^+] = 2 \times 0.05 \text{ mol dm}^{-3}$ $[H^+] = 0.1 \text{ mol dm}^{-3}$

3. Calculate the pH: $\text{pH} = -\log_{10}(0.1)$ Recall that $0.1$ can be written as $10^{-1}$. $\text{pH} = -\log_{10}(10^{-1})$ Using the logarithm rule $\log(a^b) = b \log(a)$: $\text{pH} = -(-1) \times \log_{10}(10)$ Since $\log_{10}(10) = 1$: $\text{pH} = 1 \times 1 = 1.00$

Conclusion

The calculated pH is 1.00. Comparing this to the given options, it matches option C.

Exam Tip

Always check if the acid is monoprotic (like $HCl$, which gives $1H^+$) or diprotic (like $H_2SO_4$, which gives $2H^+$). A common mistake students make is forgetting to multiply the concentration by 2 for sulfuric acid, which would lead to an incorrect answer of 1.30 (Option A). Don't fall for that trap!