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What is the pH of the solution resulting from the titration of 25 mL of 0.10 NaOH to 50.0 mL of a 0.10 M CH₃COOH? (Ka = 1.8 x 10⁻⁵)

User Pawegio
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Final answer:

At the equivalence point of a titration between 0.10 M NaOH and 0.10 M acetic acid, where equal moles of acid and base react, the solution becomes a buffer of sodium acetate and will have a pH equal to the pKa of acetic acid, which is approximately 4.74.

Step-by-step explanation:

The question pertains to the titration of acetic acid (CH₃COOH) with sodium hydroxide (NaOH). When 25.00 mL of 0.10 M NaOH is added to 50.00 mL of 0.10 M acetic acid, the moles of NaOH and acetic acid become equal, indicating the equivalence point has been reached. Since acetic acid is a weak acid and NaOH is a strong base, the resulting solution will consist of sodium acetate (CH₃COO⁻), which partially hydrolyzes in water and makes the solution weakly basic. To find the pH at the equivalence point, we use the formula pH = pKa + log([A⁻]/[HA]), where [A⁻] is the concentration of acetate ion and [HA] is the concentration of acetic acid. As all the acetic acid has been converted to acetate, the ratio [A⁻]/[HA] is 1, and the pH equals the pKa of acetic acid. To get the pKa, we use the relation pKa = -log(Ka), which gives us pKa = 4.74. Therefore, the pH of the solution at the equivalence point is approximately 4.74.

User Franz Ebner
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