What is the pH of a solution of 25.0 mL of 0.250 M HNO₃ after the addition oF₄0.0 mL of 0.300 M KOH?

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Quizplus · Accepted Answer

To find the pH of the solution after the addition of KOH to HNO3, we first need to determine if the reaction goes to completion and if there is any excess reactant. The reaction between HNO3 and KOH is a neutralization reaction, where HNO3 is a strong acid and KOH is a strong base. The balanced chemical equation for the reaction is:$$ \text{HNO}_3 + \text{KOH} \rightarrow \text{KNO}_3 + \text{H}_2\text{O} $$To find out if there's any excess reactant, we calculate the moles of HNO3 and KOH initially present:- Moles of HNO3 = 0.250 M * 0.025 L = 0.00625 mol- Moles of KOH = 0.300 M * 0.050 L = 0.015 molSince KOH is in excess (0.015 mol - 0.00625 mol = 0.00875 mol of KOH left), the solution will be basic after the reaction. The concentration of the excess KOH in the final solution (volume = 25.0 mL + 50.0 mL = 75.0 mL or 0.075 L) is:$$ \text{Concentration of KOH} = \frac{0.00875 \text{ mol}}{0.075 \text{ L}} = 0.1167 \text{ M} $$The pOH of the solution can be found using the formula:$$ \text{pOH} = -\log[\text{OH}^-] $$Given that $[OH^-] = 0.1167 \text{ M}$, we find:$$ \text{pOH} = -\log(0.1167) \approx 0.93 $$Since pH + pOH = 14 for aqueous solutions at 25°C, we find:$$ \text{pH} = 14 - \text{pOH} = 14 - 0.93 = 13.07 $$The closest answer to our calculation is C) 12.95, acknowledging a slight discrepancy due to rounding or approximation in the calculation steps.

What Is the PH of a Solution of 25