In: Chemistry
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