Question

Suppose there is 1.00 L of an aqueous buffer containing 60.0 mmol of formic acid (p?a=3.74) and 40.0 mmol of formate. Calculate the pH of this buffer. What volume of 6.00 M NaOH would be required to increase the pH to 4.93?

What volume of 6.00 M NaOH6.00 M NaOH would be required to increase the pH to 4.93?

Answer 1

Given

Volume of Buffer solution = 1 L

mmol of Formic acid = 60.0 mmol = 60.0 / 1000 = 0.06 mol

mmol of formate = 40.0 mmol = 40.0 / 1000 = 0.04 mol

[ HCOOH] = No of moles of Solute / Volume of solution in L

= 0.06 mol / 1 L

= 0.06 M

[ HCOO - ] = 0.04 mol / 1 L

= 0.04 M

pH of buffer solution calculated by using Henderson's equation

pH = pKa + log[ HCOO - ] /  [ HCOOH]

= 3.74 + log 0.04 / 0.06

= 3.74 - 0.1761

= 3.56

NaOH is a strong base . When it is added to Buffer solution it reacts with acidic component of buffer solution . In this case, NaOH reacts with HCOOH and produces sodium formate . Hence concentration of acid decreases and that of conjugate base ( formate ) increases.

NaOH + HCOOH HCOONa + H₂O

Lets use ICE table

Concentration ( moles)	HCOOH	NaOH	HCOO -
Initial	0.06	X	0.04
Change	-X	-X	+X
Equilibrium	0.06 - X	0.00	0.04 + X

Therefore, 4.93 = 3.74 + log ( 0.04 + X) / ( 0.06 - X)

log ( 0.04 + X) / ( 0.06 - X) = 4.93-3.74

= 1.19

( 0.04 + X) / ( 0.06 - X) = 10 ^1.19  = 15.5

( 0.04 + X) = 15.5 ( 0.06 - X)

= 0.93 - 15.5 X

X + 15.5 X = 0.93- 0.04 = 0.89

16.5 X = 0.89

X = 0.89 / 16.5

= 0.054 moles

We have 6.0 M NaOH .

Volume of 6.0 M NaOH required = no of moles / molarity

= 0.054 mol / 6.0 mol / L

= 0.009 L

= 9 ml

ANSWER : 9 ml of 6.0 M NaOH must be added to buffer to increase pH up to 4.93