What will be the percentage error in measuring hydrogen ion concentration in a solution on neglectin — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

If water is neglected, $[\text{H}^+]_{\text{approx}} = 1.0 \times 10^{-6}\text{ M}$. Including water, we solve $[\text{H}^+]^2 - 10^{-6}[\text{H}^+] - 10^{-14} = 0$, giving $[\text{H}^+]_{\text{exact}} = \frac{10^{-6} + \sqrt{10^{-12} + 4 \times 10^{-14}}}{2} = \frac{10^{-6}(1 + \sqrt{1.04})}{2}$. Since $\sqrt{1.04} \approx 1.0198$, $[\text{H}^+]_{\text{exact}} = \frac{10^{-6}(2.0198)}{2} = 1.0099 \times 10^{-6}\text{ M}$. The error is the difference divided by the exact value: $\frac{1.0099 - 1.000}{1.0099} \times 100\% \approx 0.98\%$.