Find the hydrogen ion, H$^+$ concentration and hydroxide ion, OH$^-$ concentration in 0.06 moldm$^{-3}$ solution of H$_2$SO$_4$.

Analysis

To solve this, we need to understand the dissociation of sulfuric acid ($H_2SO_4$) in water and the relationship between hydrogen ions ($H^+$) and hydroxide ions ($OH^-$) in an aqueous solution.

Concept

1. Dissociation: Sulfuric acid is a strong diprotic acid, meaning it dissociates completely in water according to the equation: $H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$ This means 1 mole of $H_2SO_4$ produces 2 moles of $H^+$.
2. Ionic Product of Water ($K_w$): At 25°C, the product of the concentrations of $H^+$ and $OH^-$ is constant: $K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ mol}^2\text{dm}^{-6}$

Step-by-Step Solution

Step 1: Calculate the concentration of $H^+$ ions. Since the concentration of $H_2SO_4$ is $0.06 \text{ mol dm}^{-3}$, and each molecule releases two $H^+$ ions: $[H^+] = 2 \times [\text{Concentration of } H_2SO_4]$ $[H^+] = 2 \times 0.06 = 0.12 \text{ mol dm}^{-3}$ In scientific notation, $0.12 = 1.2 \times 10^{-1} \text{ mol dm}^{-3}$.

Step 2: Calculate the concentration of $OH^-$ ions. Using the $K_w$ formula: $[OH^-] = \frac{K_w}{[H^+]}$ $[OH^-] = \frac{1.0 \times 10^{-14}}{1.2 \times 10^{-1}}$ $[OH^-] = \frac{1}{1.2} \times 10^{-13}$ $[OH^-] \approx 0.833 \times 10^{-13}$ Adjusting to standard scientific notation: $[OH^-] = 8.33 \times 10^{-14} \text{ mol dm}^{-3}$

Conclusion

Comparing our results ($[H^+] = 1.2 \times 10^{-1}$ and $[OH^-] = 8.3 \times 10^{-14}$) with the given options, we find that they match option C.

Exam Tip: Always remember that for strong acids, the concentration of $H^+$ is the molarity of the acid multiplied by its basicity (the number of replaceable hydrogen atoms). Don't forget to check if the acid is monoprotic (like $HCl$) or diprotic (like $H_2SO_4$) before calculating!