When 200. mL of 2.0 M NaOH(aq) is added to 500. mL of 1.0 M HCl(aq), the pH of the resulting mixture — Acids and Bases Chemistry Question

STEPS:

1. Calculate the initial moles of hydroxide ions ($\text{OH}^-$) added: Using the volume and molarity of the $\text{NaOH}$ solution:
$$\text{Moles of }\text{OH}^- = \text{Volume (L)} \times \text{Molarity (M)} = 0.200\text{ L} \times 2.0\text{ mol/L} = \mathbf{0.40\text{ mol}} \quad$$
2. Calculate the initial moles of hydronium ions ($\text{H}^+$) present: Using the volume and molarity of the $\text{HCl}$ solution:
$$\text{Moles of }\text{H}^+ = \text{Volume (L)} \times \text{Molarity (M)} = 0.500\text{ L} \times 1.0\text{ mol/L} = \mathbf{0.50\text{ mol}} \quad$$
3. Determine the stoichiometry of the neutralization reaction: Since $\text{HCl}$ and $\text{NaOH}$ are a strong acid and a strong base, they neutralize each other in a 1:1 mole ratio:
$$\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)$$
4. Identify the excess reactant and calculate the remaining moles: Because the moles of $\text{H}^+$ ($0.50\text{ mol}$) exceed the moles of $\text{OH}^-$ ($0.40\text{ mol}$), the $\text{OH}^-$ is the limiting reactant and is fully consumed, leaving excess acid:
$$\text{Remaining moles of }\text{H}^+ = 0.50\text{ mol} - 0.40\text{ mol} = \mathbf{0.10\text{ mol}} \quad$$
5. Determine the total volume of the mixture: Add the volumes of the two solutions together:
$$\text{Total Volume} = 200.\text{ mL} + 500.\text{ mL} = 700.\text{ mL} = \mathbf{0.700\text{ L}} \quad$$
6. Calculate the final molarity of the excess $\text{H}^+$ ions: Divide the remaining moles of $\text{H}^+$ by the total volume of the mixture:
$$[\text{H}^+] = \frac{0.10\text{ mol}}{0.700\text{ L}} = \frac{1}{7}\text{ M} \approx \mathbf{0.14\text{ M}}$$
7. Calculate the pH of the mixture:
$$\text{pH} = -\log[\text{H}^+] = -\log(0.14)$$
*Mental Math Estimation:* Since a $[\text{H}^+]$ of $0.10\text{ M}$ corresponds to a pH of $1.0$, a concentration of $0.14\text{ M}$ (which is slightly higher) will have a pH slightly lower than $1.0$ (specifically, $\text{pH} \approx 0.85$). This is closest to 1.0, which matches Option A.

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WHY_OTHERS_WRONG:

• B is incorrect: A pH of 3.0 represents a far more dilute hydronium ion concentration ($1.0 \times 10^{-3}\text{ M}$). A student might select this option if they incorrectly calculated the remaining moles of $\text{H}^+$ or made a decimal placement error during the division.
• C is incorrect: A pH of 7.0 represents a neutral solution. This would only occur if the moles of $\text{H}^+$ and $\text{OH}^-$ were stoichiometrically equal (resulting in complete neutralization). Since the acid is in excess, the solution must be strongly acidic (pH < 7).
• D is incorrect: A pH of 13.0 represents a strongly basic solution ($[\text{OH}^-] = 0.10\text{ M}$). A student might choose this option if they erroneously believed that $\text{NaOH}$ was the excess reactant rather than the limiting reactant.