A- pH = 13.12
B- pH = 12.91
C- pH = 12.71
D- pH = 12.43
E- pH = 11.55
F- pH = 7
G- pH = 2.46
H- pH = 1.88
Explanation:
This is a titration of a strong base with a strong acid. The neutralization reaction is: KOH (aq) + HCl (aq) → H₂O(l) + KCl(aq)
Our pH at the equivalence point is 7, because we have made a neutral salt.
To determine the volume at that point we state the formula for titration:
mmoles of base = mmoles of acid
Volume of base . M of base = Volume of acid . M of acid
50mL . 0.129M = 0.258 M . Volume of acid
Volume of acid = (50mL . 0.129M) / 0.258 M → 25 mL (Point F)
When we add 25 mL of HCl, our pH will be 7.
A- At 0 mL of acid, we only have base.
KOH → K⁺ + OH⁻
[OH⁻] = 0.129 M
To make more easy the operations we will use, mmol.
mol . 1000 = mmoles → mmoles / mL = M
- log 0.129 = 0.889
14 - 0.889 = 13.12
B- In this case we are adding, (7 mL . 0.258M) = 1.81 mmoles of H⁺
Initially we have 0.129 M . 50 mL = 6.45 mmoles of OH⁻
1.81 mmoles of H⁺ will neutralize, the 6.45 mmoles of OH⁻ so:
6.45 mmol - 1.81 = 4.64 mmoles of OH⁻
This mmoles of OH⁻ are not at 50 mL anymore, because our volume has changed. (Now, we have 50 mL of base + 7 mL of acid) = 57 mL of total volume.
[OH⁻] = 4.64 mmoles / 57 mL = 0.0815 M
- log 0.0815 M = 1.09 → pOH
pH = 14 - pOH → 14 - 1.09 = 12.91
C- In this case we add (12.5 mL . 0.258M) = 3.22 mmoles of H⁺
Our initial mmoles of OH⁻ would not change through all the titration.
Then 6.45 mmoles of OH⁻ are neutralized by 3.22 mmoles of H⁺.
6.45 mmoles of OH⁻ - 3.22 mmoles of H⁺ = 3.23 mmoles of OH⁻
Total volume is: 50 mL of base + 12.5 mL = 62.5 mL
[OH⁻] = 3.23 mmol / 62.5 mL = 0.0517 M
- log 0.0517 = 1.29 → pOH
14 - 1.11 = 12.71
D- We add (18 mL . 0.258M) = 4.64 mmoles of H⁺
6.45 mmoles of OH⁻ are neutralized by 4.64 mmoles of H⁺.
6.45 mmoles of OH⁻ - 4.64 mmoles of H⁺ = 1.81 mmoles of OH⁻
Total volume is: 50 mL of base + 18 mL = 68 mL
[OH⁻] = 1.81 mmol / 68 mL = 0.0265 M
- log 0.0265 = 1.57 → pOH
14 - 1.57 = 12.43
E- We add (24 mL . 0.258M) = 6.19 mmoles of H⁺
6.45 mmoles of OH⁻ are neutralized by 6.19 mmoles of H⁺.
6.45 mmoles of OH⁻ - 6.19 mmoles of H⁺ = 0.26 mmoles of OH⁻
Total volume is: 50 mL of base + 24 mL = 74 mL
[OH⁻] = 0.26 mmol / 74 mL = 3.51×10⁻³ M
- log 3.51×10⁻³ = 2.45 → pOH
14 - 2.45 = 11.55
F- This the equivalence point.
mmoles of OH⁻ = mmoles of H⁺
We add (25 mL . 0.258M) = 6.45 mmoles of H⁺
All the OH⁻ are neutralized.
OH⁻ + H⁺ ⇄ H₂O Kw
[OH⁻] = √1×10⁻¹⁴ → 1×10⁻⁷ → pOH = 7
pH → 14 - 7 = 7
G- In this case we have an excess of H⁻
We add (26 mL . 0.258M ) = 6.71 mmoles of H⁺
We neutralized all the OH⁻ but some H⁺ remain after the equilibrium
6.71 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 0.26 mmoles of H⁺
[H⁺] = 0.26 mmol / Total volume
Total volume is: 50 mL + 26 mL → 76 mL
[H⁺] = 0.26 mmol / 76 mL → 3.42×10⁻³ M
- log 3.42×10⁻³ = 2.46 → pH
H- Now we add (29 mL . 0.258M) = 7.48 mmoles of H⁺
We neutralized all the OH⁻ but some H⁺ remain after the equilibrium
7.48 mmoles of H⁺ - 6.45 mmoles of OH⁻ = 1.03 mmoles of protons
Total volume is 50 mL + 29 mL = 79 mL
[H⁺] = 1.03 mmol / 79 mL → 0.0130 M
- log 0.0130 = 1.88 → pH
After equivalence point, pH will be totally acid, because we always have an excess of protons. Before the equivalence point, pH is basic, because we still have OH⁻ and these hydroxides, will be neutralized through the titration, as we add acid.
