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Acids and Bases Calculating the \( \mathrm{pH} \) at equivalence of a titration \( 0 / 5 \) A chemist titrates \( 250.0 \mathrm{~mL} \) of a \( 0.7308 \mathrm{M} \) propionic acid \( \left(\mathrm{HC}_{2} \mathrm{H}_{5} \mathrm{CO}_{2}\right) \) solution with \( 0.5224 \mathrm{MNaOH} \) solution at \( 25^{\circ} \mathrm{C} \). Calculate the pH at equivalence. The \( p K_{a} \) of propionic acid is 4.89 . Round your answer to 2 decimal places. Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of \( \mathrm{NaOH} \) solution added. \[ \mathrm{pH}= \]

          Acids and Bases
Calculating the \( \mathrm{pH} \) at equivalence of a titration
\( 0 / 5 \)
A chemist titrates \( 250.0 \mathrm{~mL} \) of a \( 0.7308 \mathrm{M} \) propionic acid \( \left(\mathrm{HC}_{2} \mathrm{H}_{5} \mathrm{CO}_{2}\right) \) solution with \( 0.5224 \mathrm{MNaOH} \) solution at \( 25^{\circ} \mathrm{C} \). Calculate the pH at equivalence. The \( p K_{a} \) of propionic acid is 4.89 .

Round your answer to 2 decimal places.
Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of \( \mathrm{NaOH} \) solution added.
\[
\mathrm{pH}=
\]

Acids and Bases
Calculating the pH at equivalence of a titration
0 / 5
A chemist titrates 250.0 mL of a 0.7308 M propionic acid (HC2H5CO2) solution with 0.5224 MNaOH solution at 25^∘C. Calculate the pH at equivalence. The p Ka of propionic acid is 4.89 .

Round your answer to 2 decimal places.
Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of NaOH solution added.

pH=

Added by Shania E.

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