Question

You need to prepare an acetate buffer of pH of 6.39 from a 0.876 M acetic acid solution and a 2.52 M KOH solution. If you have 775mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a buffer of pH 6.39? The pKa of acetic acid is 4.76.

Answer 1

Let's write down the chemical equation for the reaction that occurs between acetic acid and KOH

CH3COOH + KOH -------------> CH3COOK + H2O

here,CH3COOH is a weak acid and CH3COOK is its strong salt . Together they make up the buffering system known as acetate buffer.

Let's find equilibrium concentrations of CH3COOH and CH3COOK

Initial moles of CH3COOH = conc. of acid * liters = 0.876 mol/L * 775 mL * 1 L / 1000 mL = 0.6789 mol

Let's say we are adding x moles of KOH

	CH3COOH	KOH	CH3COOK	H2O
I	0.6789	x	0	-
C	-x	-x	+x	-
E	0.6789 - x	0	x	-

At equilibrium we can see that we have only weak acid CH3COOH and its conjugate base CH3COOK .

Let's use Henderson Hasselbalch equation

The equation is written as

pH = pKa + log [ moles of base / moles of acid ] ............Note : we can use both moles and concentration terms because its a ratio.

Here, pH is pH of the buffer which is given as 6.39

pKa is 4.76

moles of acid = 0.6789 - x

moles of conjugate base = x

Let's plug in above values

6.39 = 4.76 + log [ x / 0.6789 - x ]

1.63 = log [ x / 0.6789 - x ]

10^1.63 = [ x / 0.6789 - x ]

x / 0.6789 - x = 42.66

x = 42.66 ( 0.6789 - x )

x = 28.96 - 42.66 x

43.66 x = 28.96

x = 0.663 mol

But x is moles of CH3COOK , and from the neutralization equation, we know that stoichiometry for KOH and CH3COOK is 1:1

Therefore we can say that moles of KOH added = 0.663

Molarity of KOH = moles of KOH / liters

2.52 M = 0.663 mol/ Liters

Liters of KOH = 0.663 / 2.52 = 0.263

0.263 L = 263 mL

We need 263 mL of base to make the given buffer