Question

# A buffer is prepared by taking 250.0mL of a 0.0865 M acetic acid solution, adding 1.232...

A buffer is prepared by taking 250.0mL of a 0.0865 M acetic acid solution, adding 1.232 grams of potassium hydroxide to the solution, then diluting the solution to 500.00mL. what is the pH of the buffer solution?

#### Homework Answers

Answer #1

Given,

Molarity of Acetic Acid solution = 0.0865 M

Volume = 250 mL = 0.25 L

=> Moles of Acetic Acid = 0.0865 x 0.25 = 0.021625 moles

Mass of KOH = 1.232 g

=> Moles of KOH = 1.232 / 56.1 = 0.022 moles

The reaction is,

CH3COOH + KOH -------> CH3COO- + H2O + K+

According to the stoichiometry of the reaction 1 mole of CH3COOH reacts with 1 mole of KOH

=> 0.021625 moles of CH3COOH reacts with 0.021625 moles of KOH

=> Moles of KOH left = 0.022 - 0.021625 = 3.75 x 10^-4 moles

Volume of solution = 0.25 L

=> [KOH] = 3.75 x 10^-4 / 0.25 = 1.5 x 10^-3 M

Since KOH is a strong base it dissociates completely as follows

KOH -----> K+ + OH-

=> [OH-] = 1.5 x 10^-3 M

We know that,

pOH = - log [OH-] = - log (1.5 x 10^-3) = 2.824

pH = 14 - pOH = 14 - 2.824 = 11.176

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