A 20.0-mL sample of 0.25 M HNO₃ is titrated with 0.15 M NaOH. What is the pH of the solution after 30.0 mL of NaOH have been added to the acid?

Question

Quizplus · Accepted Answer

The reaction between HNO3 (a strong acid) and NaOH (a strong base) is a neutralization reaction, where 1 mole of HNO3 reacts with 1 mole of NaOH to produce NaNO3 and H2O. The moles of HNO3 initially present are $20.0 \, \text{mL} \times 0.25 \, \text{M} = 5.0 \, \text{mmol}$. The moles of NaOH added are $30.0 \, \text{mL} \times 0.15 \, \text{M} = 4.5 \, \text{mmol}$. After the reaction, there are $5.0 \, \text{mmol} - 4.5 \, \text{mmol} = 0.5 \, \text{mmol}$ of HNO3 left unreacted in a total volume of $20.0 \, \text{mL} + 30.0 \, \text{mL} = 50.0 \, \text{mL}$. The concentration of HNO3 is then $0.5 \, \text{mmol} / 50.0 \, \text{mL} = 0.01 \, \text{M}$. Since HNO3 is a strong acid, its pH can be calculated directly from its concentration: $\text{pH} = -\log[0.01] = 2.00$.

A 200-ML Sample of 0