Question

Calculate the pH of the solution that results from mixing 35.0 mL of 0.18 M formic acid and 70.0 mL of 0.090 M sodium hydroxide. ( value for formic acid is 1.8*10-4.)

Answer 1

Given:

M(HCOOH) = 0.18 M

V(HCOOH) = 35 mL

M(NaOH) = 0.09 M

V(NaOH) = 70 mL

mol(HCOOH) = M(HCOOH) * V(HCOOH)

mol(HCOOH) = 0.18 M * 35 mL = 6.3 mmol

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.09 M * 70 mL = 6.3 mmol

We have:

mol(HCOOH) = 6.3 mmol

mol(NaOH) = 6.3 mmol

6.3 mmol of both will react to form HCOO- and H2O

HCOO- here is strong base

HCOO- formed = 6.3 mmol

Volume of Solution = 35 + 70 = 105 mL

Kb of HCOO- = Kw/Ka = 1*10^-14/1.8*10^-4 = 5.556*10^-11

concentration ofHCOO-,c = 6.3 mmol/105 mL = 0.06M

HCOO- dissociates as

HCOO- + H2O -----> HCOOH + OH-

0.06 0 0

0.06-x x x

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

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.556*10^-11)*6*10^-2) = 1.826*10^-6

since c is much greater than x, our assumption is correct

so, x = 1.826*10^-6 M

[OH-] = x = 1.826*10^-6 M

use:

pOH = -log [OH-]

= -log (1.826*10^-6)

= 5.7386

use:

PH = 14 - pOH

= 14 - 5.7386

= 8.2614

Answer: 8.26