A 20.0-mL sample of 0.30 M HBr is titrated with 0.15 M NaOH. What is the pH of the solution after 40.3 mL of NaOH have been added to the acid?
A)2.95
B)3.13
C)10.87
D)11.05
E)13.14

Question

Quizplus · Accepted Answer

The reaction between HBr and NaOH is a neutralization reaction, where HBr (a strong acid) reacts with NaOH (a strong base) to form water and NaBr (a salt). The stoichiometry of the reaction is 1:1. Initially, there are $20.0 \, \text{mL} \times 0.30 \, \text{M} = 6.0 \, \text{mmol}$ of HBr. $40.3 \, \text{mL} \times 0.15 \, \text{M} = 6.045 \, \text{mmol}$ of NaOH are added, which is slightly more than the amount of HBr. This means all the HBr will react, and there will be a small excess of NaOH. The excess NaOH is $6.045 \, \text{mmol} - 6.0 \, \text{mmol} = 0.045 \, \text{mmol}$. The total volume after mixing is $20.0 \, \text{mL} + 40.3 \, \text{mL} = 60.3 \, \text{mL}$. The concentration of the excess NaOH is $0.045 \, \text{mmol} / 60.3 \, \text{mL} = 0.000746 \, \text{M}$. The pH of a solution of NaOH at this concentration can be found by calculating the pOH first: $pOH = -\log[OH^-] = -\log(0.000746)$, and then converting to pH: $pH = 14 - pOH$. This calculation gives a pH close to 10.87, indicating the solution is basic due to the excess NaOH.

A 200-ML Sample of 0