\begin{tabular}{l} 【3-51 1] 1 L of HCl acid its \( p^{\mathrm{H}} \) value =1 was added to 1L \\ of NaOH its \( p^{\mathrm{H}} \) value \( =11 \). \\ What is \( p^{\mathrm{H}} \) value of that mixture? \\ a) 0.996 \\ \( \begin{array}{l}\text { b) } 2 \\ \text { c) } 1.3 \\ \text { d) } 1.703 \\ \text { Type here to search }\end{array} \) \\ \hline\end{tabular}

To determine the \( \mathrm{pH} \) of the mixture when 1 L of HCl (with \( \mathrm{pH} = 1 \)) is mixed with 1 L of NaOH (with \( \mathrm{pH} = 11 \)), follow these steps: 1. **Calculate the Concentrations:** - **HCl Solution:** \[ \mathrm{pH} = 1 \implies [\mathrm{H}^+] = 10^{-1} \, \text{M} = 0.1 \, \text{M} \] Volume = 1 L \(\Rightarrow\) Moles of H\(^+\) = \(0.1 \times 1 = 0.1 \, \text{mol}\) - **NaOH Solution:** \[ \mathrm{pH} = 11 \implies \mathrm{pOH} = 14 - 11 = 3 \implies [\mathrm{OH}^-] = 10^{-3} \, \text{M} = 0.001 \, \text{M} \] Volume = 1 L \(\Rightarrow\) Moles of OH\(^-\) = \(0.001 \times 1 = 0.001 \, \text{mol}\) 2. **Neutralization Reaction:** \[ \mathrm{H}^+ + \mathrm{OH}^- \rightarrow \mathrm{H}_2\mathrm{O} \] - Moles of H\(^+\) remaining after reaction: \[ 0.1 \, \text{mol} - 0.001 \, \text{mol} = 0.099 \, \text{mol} \] 3. **Determine the Final Concentration:** - Total volume after mixing = \(1 \, \text{L} + 1 \, \text{L} = 2 \, \text{L}\) - Concentration of H\(^+\): \[ [\mathrm{H}^+] = \frac{0.099 \, \text{mol}}{2 \, \text{L}} = 0.0495 \, \text{M} \] 4. **Calculate the \( \mathrm{pH} \):** \[ \mathrm{pH} = -\log(0.0495) \approx 1.3 \] **Answer:** **c) 1.3**