Question

Prepare a buffer by acid–base reactions.

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Consider how to prepare a buffer solution with pH = 9.31 (using one of the weak acid/conjugate base systems shown here) by combining 1.00 L of a 0.316-M solution of weak acid with 0.351 M potassium hydroxide.

Weak Acid	Conjugate Base	Ka	pKa
HNO2	NO2-	4.5 x 10-4	3.35
HClO	ClO-	3.5 x 10-8	7.46
HCN	CN-	4.0 x 10-10	9.40

How many L of the potassium hydroxide solution would have to be added to the acid solution of your choice?

L

Answer 1

We will choose HCN as the weak acid since its pKa value is much closer to the pH required.

pKa for KCN = 9.40

Conc. of Acid added = 0.316 M

Volume of Acid added = 1 L

=> Moles of Acid added = 0.316 x 1 = 0.316 moles

The reaction is,

HCN + KOH ----> KCN + H2O

pH of the buffer = pKa + log (KCN / HCN)

=> 9.31 = 9.40 + log (KCN / HCN)

=> log (KCN / HCN) = - 0.09

=> (KCN / HCN) = 0.813

=> [KCN] = 0.813 * [HCN]

Suppose we add 'n' moles of KOH

HCN + KOH ----> KCN + H2O

1 - n .....n - n .........n ........n

=> n / 1-n = 0.813

=> n = 0.4484 moles of KOH

Molarity of KOH = 0.351 M

Volume = Moles / Molarity = 0.4484 / 0.351 = 1.278 liters