Question

17.2c Making Buffer Solutions:

Consider how to prepare a buffer solution with pH = 9.24 (using one of the weak acid/conjugate base systems shown here) by combining 1.00 L of a 0.330-M solution of weak acid with 0.215 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

Sol .

Weak acid choosen is HCN . As pKa of HCN is 9.40 which is so close to pH of buffer solution , 9.24 .

Now , initial Moles of weak acid , HCN = Conc. of HCN × Volume of HCN = 0.330 × 1 = 0.330 mol

Let volume of KOH (potassium hydroxide ) added = x  

So , Moles of KOH added = Conc. of KOH × Volume of KOH added = 0.215 x mol

Now , Reaction :

HCN + OH- <----> CN- + H2O

initial 0.330 0.215x 0

change - 0.215x -0.215x + 0.215x

equilibrium (0.330- 0.215x) 0 0.215x

Also , pKa = 9.40

So , Using Henderson - Hasselbalch equation ,

pH = pKa + log ( Equilibrium moles of CN- / Equilibrium moles of HCN )

9.24 = 9.40 + log ( 0.215x / (0.330 - 0.215x ) )

0.215x / (0.330 - 0.215x) = 10^-0.16   = 0.6918

0.215x = 0.6918 ( 0.330 - 0.215x)

0.215x = 0.228294 - 0.148737x

x = 0.627  

Therefore , Volume of KOH added =  0.627 L