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?

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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