Question

Calculate the pH at the stoichiometric point when 25 mL of 0.089M formic acid is titrated with 0.33 M NaOH.

Calculate the pH at the stoichiometric point when 25 mL of 0.096 M pyridine is titrated with 0.31 M HCl.

Calculate the pH at the stoichiometric point when 25 mL of 0.093 M nitric acid is titrated with 0.34 M NaOH.

Answer 1

1 )

25 mL of 0.089M formic acid is titrated with 0.33 M NaOH.

pKa of formic acid = 3.75

At the stoichiometric point

V1x M1 = v2 x M2

25mL x 0.089 M = 0.33M x V2

Volume of NaOH = 6.74 mL

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

25x0.089 6.74 x0.33 0 0- intial mmoles

00000 0000 25x0.089 ------ at equivalence

[salt ] formed = mmoles/ volume = 25x0.089/(25+6.74) =0.070 M

pH of salt of weak acid and strong base is given by

pH = 1/2[pKw +pKa + logC]

= 1/2[ 14+ 3.75 +log 0.070} =8.297

2)25 mL of 0.096 M pyridine is titrated with 0.31 M HCl.

pKb of py = 8.77

At equivalence

V1M1 = V2M2

25x0.096 = V x0.31

V of HCl = 7.74

and the salt formed has molarity = mmoles/ total volume = 25x0.096 /(25+7.74) =0.0733

pH of salt = 1/2[ pKw -pKb -log C]

= 1/2[ 14-8.77-log 0.0733] =3.18

3) 25 mL of 0.093 M nitric acid is titrated with 0.34 M NaOH.

As both NaOH and nitric acid are strong, the salt formed does not undergo hydrolysis and hence is neutral pH = 7.0