Consider the titration of a 25.0 mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH.
ID: 1030088 • Letter: C
Question
Consider the titration of a 25.0 mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. Determine each of the following.
(1.)
the initial pH
Express your answer using two decimal places.
(2.)
the volume of added base required to reach the equivalence point
(3.)
the pH at 5.00 mL of added base
Express your answer using two decimal places.
(4.)
the pH at one-half of the equivalence point
Express your answer using two decimal place
(5.)
the pH at the equivalence point
Express your answer using two decimal places.
Explanation / Answer
1) You have a weak acid. In order to calculate pH you first have to determine the [H+] in the solution. You do this using the Ka equation:
Ka = [H+] [CH3COO-] / [CH3COOH]
You know that [H+] = [CH3COO-] so for product we write [H+]²
Because the acid is weak , we say [CH3COOH] = 0.10
Substitute:
1.8*10^-5 = [H+]² / 0.11
[H+]² = (1.8*10^-5) *0.1
[H+]² = 1.98*10^-6
[H+] = (1.98*10^-6)
[H+] = 1.41*10^-3
pH = -log [H+]
pH = -log (1.41*10^-3)
pH = 2.85
2)
At equivalence point the number of moles of acid = number of moles of base.
Moles of CH3COOH = 0.110 M x 0.025 L = 0.00275 moles
So, the moles of NaOH = 0.00275 mol
Molarity of NaOH = 0.130 M
So, the volume of NaOH = (0.00275 mol) / 0.130 M
= 0.02115 L
= 21 mL
3)
Moles of CH3COOH = 0.110 M x 0.025 L = 0.00275 moles
Moles of NaOH = 0.130 M x 0.005 L = 0.00065 moles
When you have added 5 mL of 0.130 M NaOH solution, it will react with CH3COOH to produce CH3COONa. So, 0.00065 moles of NaOH will react with 0.00065 moles of CH3COOH to produce 0.00065 moles of CH3COONa.
Hence, moles of CH3COOH remain = 0.00275 mol - 0.00065 mol = 0.0021 mol
Total volume = 25 mL + 5 mL = 30 mL = 0.030 L
Hence, the concentration terms are
[CH3COOH] = (0.0021 mol) / (0.030 L) = 0.07 M
[CH3COONa] = (0.00065 mol) / (0.030 L) = 0.022 M
Ka of CH3COOH = 1.8 * 10^-5
pKa = - log Ka
= - log (1.8 * 10^-5)
= 4.74
using the Henderson - Hasselbalch equation.
pH = pKa + log ([salt] /[acid])
pH = pKa + log ([CH3COONa] / [CH3COOH])
pH = 4.74 + log (0.022 / 0.07)
pH = 4.74 + log (0.31)
PH = 4.74 - 0.51
pH = 4.23
5)
At the equivalence point; Moles of acid = Moles of base. So, when you have added 21 mL of 0.130 M NaOH solution you have neutralized all the acid. There is then only a solution of salt CH3COONa in solution
0.00275 moles of CH3COOH reacts with 0.00275 moles of NaOH to produce 0.00275 moles of CH3COONa.
Total volume = 25 mL + 21 mL
= 46 mL
= 0.046 L
So, [CH3COONa] = 0.00275 mol / 0.046 L = 0.06 M
Calculate pH as follows:
The aim is to find the [OH-] of the solution. You will know that a salt of a strong base and weak acid is basic.
The CH3COONa dissociates in water :
CH3COONa CH3COO- + Na+ The CH3COO- reacts with water:
CH3COO- + H2O CH3COOH + OH-
Ka for CH3COOH = 1.8*10^-5
Kw = Ka * Kb - we want Kb
Kb = Kw* Ka
Kb = (1.0*10^-14) / (1.8*10^-5)
Kb = 5.56*10^-10
Kb = [CH3COO-] *[[OH] / [CH3COONa]
We know that [CH3COO-] = [OH-] so product is [OH-]²
[CH3COONa] = 0.05M
Kb = [OH]²/0.06
(5.56*10^-10 ) * 0.05 = [OH]²
[OH]² = 2.78*10^-12
[OH-] = 5.27*10^-6
pOH = - log [OH-]
= - log (5.27*10^-6)
= 5.28
pH = 14 - pOH
pH = 14 – 5.28
pH = 8.72