The pH of a solution made by combining 150.0 mL of 0.10 M KOH(aq) with 50.0 mL of 0.20 M HBr(aq) is — Acids and Bases Chemistry Question

STEPS:

1. Identify the nature of the reactants:
* Potassium hydroxide ($\text{KOH}$) is a strong base that dissociates completely in water to yield hydroxide ions ($\text{OH}^-$).
* Hydrobromic acid ($\text{HBr}$) is a strong acid (with a very large dissociation constant, $K_a \gg 1$) that dissociates completely to yield hydrogen ions ($\text{H}^+$).
* Because both are strong species, they react completely in a $1:1$ ratio to undergo a neutralization reaction:
$$\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \quad$$

2. Calculate the initial moles of $\text{OH}^-$ and $\text{H}^+$ ions:
Using the formula $\text{moles} = \text{Volume (L)} \times \text{Molarity}$:
* $\text{Initial moles of OH}^- = 0.1500\text{ L} \times 0.10\text{ M} = \mathbf{0.0150\text{ mol}} \quad$
* $\text{Initial moles of H}^+ = 0.0500\text{ L} \times 0.20\text{ M} = \mathbf{0.0100\text{ mol}} \quad$

3. Determine the limiting reactant and excess reactant moles:
* $\text{H}^+$ is the limiting reactant because there are fewer moles of it ($0.0100\text{ mol} < 0.0150\text{ mol}$).
* During the neutralization reaction, all $0.0100\text{ mol}$ of $\text{H}^+$ is consumed by reacting with $0.0100\text{ mol}$ of $\text{OH}^-$.
* The moles of excess $\text{OH}^-$ remaining in the solution are:
$$\text{Moles of excess OH}^- = 0.0150\text{ mol} - 0.0100\text{ mol} = \mathbf{0.0050\text{ mol}}$$

4. Calculate the total volume of the combined mixture:
* Combined Volume = $150.0\text{ mL} + 50.0\text{ mL} = 200.0\text{ mL} = \mathbf{0.2000\text{ L}} \quad$

5. Determine the equilibrium concentration of hydroxide ions, $[\text{OH}^-]$:
Divide the excess moles of hydroxide by the new total volume of the solution:
$$[\text{OH}^-] = \frac{0.0050\text{ mol}}{0.2000\text{ L}} = \mathbf{0.025\text{ M}}$$

6. Calculate pOH and convert to pH:
* On Section I of the AP Exam, calculators are not allowed. However, you can easily approximate this logarithm.
* Since $[\text{OH}^-] = 0.025\text{ M}$ lies between $0.01\text{ M}$ ($10^{-2}\text{ M}$) and $0.1\text{ M}$ ($10^{-1}\text{ M}$), its $\text{pOH}$ must fall between $1.0$ and $2.0$. Specifically, $\text{pOH} = -\log(0.025) \approx 1.6$.
* Convert $\text{pOH}$ to $\text{pH}$ using the water equilibrium relationship:
$$\text{pH} = 14 - \text{pOH} \quad$$
$$\text{pH} \approx 14 - 1.6 = \mathbf{12.4}$$
* Looking at the options, $12.4$ is closest to $12$, identifying Option D as the correct choice.

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WHY_OTHONG_WRONG:

• Option A is incorrect (2): A student might arrive at this value if they successfully calculated the excess mole value ($0.0050\text{ mol}$) but misidentified it as excess *acid* ($\text{H}^+$) rather than excess base ($\text{OH}^-$). This would lead them to calculate $[\text{H}^+] = 0.025\text{ M}$, which yields a $\text{pH}$ of $\approx 1.6$ (closest to $2$). Alternatively, they may have calculated the correct $\text{pOH}$ of $\approx 1.6 \approx 2$ and mistakenly selected it as the $\text{pH}$.
• Option B is incorrect (4): This is a distractor value that could result from severe decimal or mathematical placement errors during the calculation of moles (e.g., miscalculating excess concentration as $10^{-4}\text{ M}$), or by confusing the chemical equations with other multi-step polyprotic acid ionizations on the exam.
• Option C is incorrect (7): A student might select this value by quickly identifying that a strong base is being mixed with a strong acid and assuming that all strong acid-strong base mixtures automatically result in a perfectly neutral solution ($\text{pH} = 7$). However, this only occurs when equimolar amounts of $\text{H}^+$ and $\text{OH}^-$ react. Here, the base is in excess, so the final solution must be basic ($\text{pH} > 7$).