A 25.0 mL portion of 0.0500 M HCl was added to 10.0 mL of 0.0250 M NaOH. Calculate the pH of the resulting solution.

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Quizplus · Accepted Answer

The reaction between HCl and NaOH is a neutralization reaction where HCl (a strong acid) reacts with NaOH (a strong base) to form water and NaCl. The moles of HCl initially are $25.0 \, \text{mL} \times 0.0500 \, \text{M} = 1.25 \times 10^{-3} \, \text{moles}$, and the moles of NaOH are $10.0 \, \text{mL} \times 0.0250 \, \text{M} = 0.25 \times 10^{-3} \, \text{moles}$. After reaction, the remaining moles of HCl are $1.25 \times 10^{-3} - 0.25 \times 10^{-3} = 1.0 \times 10^{-3} \, \text{moles}$. The total volume of the solution is $25.0 \, \text{mL} + 10.0 \, \text{mL} = 35.0 \, \text{mL}$ or $0.0350 \, \text{L}$. The concentration of the remaining HCl is $\frac{1.0 \times 10^{-3} \, \text{moles}}{0.0350 \, \text{L}} = 0.0286 \, \text{M}$. The pH is calculated using the formula $\text{pH} = -\log[H^+]$, where $[H^+] = 0.0286 \, \text{M}$. Thus, $\text{pH} = -\log(0.0286) \approx 1.54$.

A 250 ML Portion of 0