Question

A buffer contains 0.010 mol of lactic acid and 0.050 mol of sodium lactate per liter. (a) calculate the pH of the buffer. (b) calculate the pH when 100ml of 0.02 M NaOH is added to 500mL of the buffer. (c) What pH would you expect if you added the same quantity of NaOH to 500mL of pure water?

Answer 1

pH of acid buffer using Hendersson-Hasselbalch equstion is

pH =pKa + log(moles of sodium lactate)/(moles of lactic acid)

pKa of lactic acid = 3.86

So, pH = 3.86+ log(0.050/0.010)

Or, pH = 3.86 + 0.698= 4.558.

b)

When NaOH is added it reacts with lactic acid (HA) to form sodium lactate (NaA).

moles of NaOH = (100*0.02)/1000 = 0.002

BCA table is

	HA	NaOH	NaA
Before	0.010	0.002	0.050
Change	-0.002	-0.002	+0.002
After	0.008	0	0.052

Now new pH is

pH = pKa + log(0.052/0.008)

= 3.86 + 0.813

= 4.673.

C)

Moles of NaOH = 100*0.02 = 0.002

When added to 500 mL water

New volume of solution = ( 500 +100 ) = 600 mL = 0.6 L.

Molar concentration of NaOH = (0.002/0.6) = 0.00333 M

Now, pOH = - log[OH^-] = - log (0.00333) = 2.48

Now expected pH of the solution will be

Or, pH = 14 - 2.48 = 11.52.