Question

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Calculate the pH at the equivalence point for the following titration: 0.40 M HCOOH versus 0.40 M NaOH.

Answer 1

firstly we have to consider according to the question the answer of this question is 14

The equation for the neutralization reaction between formic acid (HCOOH), a weak acid, and sodium hydroxide (NaOH), a strong base, is

HCOOH(aq) + NaOH(aq) <=> NaCOOH(aq) + H2O, Ka = 1.7 x 10-4

Suppose we combine equal volumes of 0.40 M HCOOH and 0.40 M NaOH. The molarities of HCOOH and NaOH in the combined volume of solution are both 0.20 M. There are sufficient amounts of H+ from HCOOH and OH- from NaOH to cause neutralization, but since HCOOH is a weak acid, the formate ion (HCOO-) undergoes hydrolysis according to

HCOO-(aq) + H2O <=> HCOOH(aq) + OH-,

Kh = Kw/Ka

= (10-14) / (1.7 x 10-4)

= 5.882 x 10-11

and the final solution will not be exactly neutral. Let x be the moles per liter of HCOO- which hydrolyze. At equilibrium, we have

[HCOO-] = 0.20 - x
[OH -] = x
[HCOOH] = x

The hydrolysis constant

Kh = [HCOOH][OH -]/[HCOO-]

= x²/(0.20 - x)

=> 0.20K_h - K_hx = x²

=> x² + K_hx - 0.20K_h = 0
=> x = {- K_h ± sqrt[Kh2 + (4)(0.20)K_h)]} / 2

=> {- K_h + sqrt[K²_h + (4)(0.20)K_h)]} / 2

= 1.079 M

Thus, [OH -] = x

=1.079 M and it is approximately 1.0 M

and[H3O+] = Kw/[OH -]

= (10^-14) / ( 1.0)

= 10^-14 M

=> pH = - log [H3O+]

= -log[10^-14]

=14