Prepare a buffer by acid–base reactions. Close Problem Consider how to prepare a
ID: 949045 • Letter: P
Question
Prepare a buffer by acid–base reactions.
Close Problem
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.
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
Weak Acid Conjugate Base Ka pKaHNO2
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
Explanation / Answer
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