500.0 mL of 0.160 M NaOH is added to 595 mL of 0.250 M weak acid (Ka = [tex]$1.69 \times 10^{-5} \, M$[/tex]). What is the pH of the resulting buffer?

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19-01-2024
Chemistry

High School

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500.0 mL of 0.160 M NaOH is added to 595 mL of 0.250 M weak acid (Ka = [tex]$1.69 \times 10^{-5} \, M$[/tex]). What is the pH of the resulting buffer?

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professorenglish94 professorenglish94 · Answer 1 · 2025-04-06T21:25:36-04:00

The pH of the resulting buffer is approximately 4.62.

To find the pH of the resulting buffer, first, we need to determine the moles of weak acid and NaOH. The moles of NaOH can be calculated by multiplying its molarity (0.160 M) by its volume (500.0 ml), resulting in 0.080 moles. The weak acid's moles can be found similarly by multiplying its molarity (0.250 M) by its volume (595 ml), which equals 0.14875 moles. Since NaOH completely reacts with the weak acid in a 1:1 ratio, the remaining excess moles of weak acid after the reaction will be 0.14875 - 0.080 = 0.06875 moles.

Next, we need to calculate the concentration of the weak acid and its conjugate base. The total volume of the resulting solution is 500.0 ml + 595 ml = 1095 ml = 1.095 L. The concentration of the weak acid is 0.14875 moles / 1.095 L ≈ 0.136 M. Since NaOH reacts with the weak acid to form its conjugate base, the concentration of the conjugate base will be 0.080 moles / 1.095 L ≈ 0.073 M.

Finally, we use the Henderson-Hasselbalch equation to find the pH of the buffer, pH = pKa + log([conjugate base]/[weak acid]). Plugging in the values, we get pH = -log(1.69 x 10^-5) + log(0.073/0.136). After calculating, we find pH ≈ 4.62. Hence, the correct option is approximately 4.62.