Question

Formic acid is a weak acid with pKa = 3.75. How many mL of 0.10 M NaOH solution should be added to 100.0 mL of 0.10 M formic acid solution to make a buffer solution with pH = 3.27

A) 0

B) 25

C) 50

D) 75

E) 100

Answer 1

Henderson- Hasselbach equation is

pH= pKa+ log [A-]/[HA]

where [HA] = concentration of formic acid and [A-]= concentration of conjugate base

moles of formic acid in 100ml of 0.1M= Molarity* volume in L= 0.1*100/1000 =0.01 moles

let x= volume of NaOH in L, hence moles of NaOH added =x*0.1= 0.1x

the reaction between formic acid and NaOH is

HCOOH+ NaOH---------->HCOONa+ H2O

here NaOH if not in excess lead to pH rise to basic value. since pH required is 3.27, limiting reagant is NaOH.

hence mole of HCOONa= 0.1x, moles of HCOOH remaining after reaction = 0.01-0.1x

volume of solution after mixing = 0.1+x L

concentrations after mixing HCOO- = 0.1x/(0.1+x) , HCOOH= (0.01-0.1x)/(x+0.1)

hence pH= Pka+ log {[HCOO-]/[HCOOH]}

3.27=3.75 + log (0.1x/(0.01-0.1x)

0.33= 0.1x/(0.01-0.1x)

0.33*(0.01-0.1x)= 0.1x

0.33*0.01= 0.1x*(1+0.33)

0.33*0.01= 0.1x*1.33

x=0.0248L, 0.0248*1000ml= 24.8 ml ( close to 25 ml). Hence B is correct.