Determine the ph during the titration of 25.0 mL of 0.320 M formic acid (Ka= 1.8

Question

Determine the ph during the titration of 25.0 mL of 0.320 M formic acid (Ka= 1.8x10^-4) by 0.356 M NaOh at the following points: 1. before the addition of NaOH 2. After the addition of 5.70 mL of NaOH 3.At the half equivalence point 4. at the equivalence point 5. after the addition of 33.7 mL of NaOH Determine the ph during the titration of 25.0 mL of 0.320 M formic acid (Ka= 1.8x10^-4) by 0.356 M NaOh at the following points: 1. before the addition of NaOH 2. After the addition of 5.70 mL of NaOH 3.At the half equivalence point 4. at the equivalence point 5. after the addition of 33.7 mL of NaOH 1. before the addition of NaOH 2. After the addition of 5.70 mL of NaOH 3.At the half equivalence point 4. at the equivalence point 5. after the addition of 33.7 mL of NaOH

Explanation / Answer

1) Since no base has been added, the pH of solution is based on the ionization of acid.

HCOOH + H2O -----> HCOO– + H3O+

Ka= 1.8x10^-4

pKa = - log 1.8x10^-4 = 3.74

Ka = [HCOO– ] [H3O+]/ [HCOOH]

1.8x10^-4 = [HCOO– ] [H3O+]/0.320

[HCOO– ] = [H3O+] = x

x^2 = 1.8 x10^-4 x 0.320

      = 5.76 x10^-5

x = 7.59 x 10^-3 = [H3O+]

pH = - log [H3O+]

pH = -log(7.59 x 10^-3)

pH = 2.12

2. after addition of 5.70 mL of NaOH

Moles of acid = (25.0 /1000) L x 0.320 M = 0.008 moles

Moles of NaOH = (5.70 /1000) L x 0.356M = 0.002 moles

0.002 moles of NaOH neutralizes 0.002 moles of acid

So, remaining moles of acid =0.008 - 0.002 = 0.006 moles

Final volume of solution = 25 + 5.5 = 30.5 ml

[HCOOH] = 0.006/(30.5/1000) = 0.1967M

[HCOO– ] = 0.002/(30.5/1000) = 0.0656 M

By using Henderson–Hassel Balch equation, we can calculate pH

pH = pKa + log [HCOO– ] /[HCOOH]

pH = 3.74 + log (0.0656/0.1967)

pH = 3.74 - 0.48 = 3.26

3. At the half equivalence point

At half equivalence point, the concentrations of the formic acid and its conjugate base are equal.

So, [HCOO– ] /[HCOOH] = 1

The log of 1 is zero, so, the pH = pKa = 3.74

pH = 3.74

At this point all the acid is neutralized by the base to produce of salt. Since only salt is present, the pH of the solution is based on hydrolysis of this salt.

[HCOO– ] = 0.008/(30.5/1000) = 0.2623 M

HCOO– + H2O ---> HCOOH +     OH-

I 0.2623                                  0                     0

C –x                          +x                    +x

E 0.2623 –x              x                     x

Kb = 1.0 x 10-14/1.8 x10^-4 = 5.66 x 10^-11

Kb = [HCOO– ] [OH-]/ [HCOOH]

5.66 x 10^-11 = [HCOO– ] [OH-]/ 0.2623 –x

x^2 = 5.66 x 10^-11 x 0.2623

x = 3.85 x 10^-6 = OH-

pOH = - log (3.85 x 10^-6) = 5.41

pH = 14 – pOH = 14 - 5.41 = 8.59

pH = 8.59

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