Respuesta :
Answer:
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.
The pH value is given by 14 less the logarithm of the hydronium ion
concentration or the hydrogen ion concentration in the solution.
Responses (approximate values):
The pH values are;
A) 13.11
B) 12.91
C) 12.71
D) 12.42
E) 11.54
F) 7
G) 2.469
H) 1.884
Which is used to find the pH of the solution?
A) Concentration of the KOH = 0.129 M
Amount of HCl added = 0.00 ml
The pH = -log[H⁺] = 14 - pOH
pOH = -log[OH⁻]
Which gives;
pH = 14 - (-log[OH⁻] )
- pH = 14 - (-log(0.129)) ≈ 13.11
B) Volume of acid added = 7.00 mL = 0.007 L
Concentration of the acid = 0.258 M HCl
Number of moles of acid, H⁺ = 0.007 × 0.258 moles = 0.001806 moles
Number of moles of KOH remaining, OH⁻= 0.05 × 0.129 - 0.001806 = 0.004644
Number of moles of OH⁻ = 0.004644 moles
[tex]Concentration, \ [OH^-] = \dfrac{0.004644 \, moles}{0.05 7 \, L} \approx \mathbf{ 0.0815 \, M}[/tex]
- pH of solution = 14 - (-log(0.0815)) ≈ 12.91
C) 12.5 mL HCl contains, 0.0125 × 0.258 moles = 0.003225 moles
OH⁻ remaining = 0.05 × 0.129 - 0.003225 = 0.003225 moles
[tex]Concentration \ of \ [OH^-]= \dfrac{0.003225\, moles}{(0.05 + 0.0125) \, L} = \mathbf{0.0516 \, M}[/tex]
- pH of solution = 14 - (-log(0.0516)) ≈ 12.71
D) 18.0 mL HCl contains, 0.018 × 0.258 moles = 0.004644 moles
OH⁻ remaining = 0.05 × 0.129 - 0.004644 = 0.001806 moles
[tex]Concentration \ of \ [OH^-] = \dfrac{0.001806\, moles}{(0.05 + 0.018) \, L} \approx \mathbf{0.0266\, M}[/tex]
- pH of solution = 14 - (-log(0.0266)) ≈ 12.42
E) 24.0 mL HCl contains, 0.024 × 0.258 moles = 0.006192 moles
OH⁻ ion remaining = 0.05 × 0.129 - 0.006192 = 0.000258 moles
[tex]Concentration \ of \ [OH^-] = \dfrac{0.000258\, moles}{(0.05 + 0.024) \, L} \approx \mathbf{0.0035\, M}[/tex]
- pH of solution = 14 - (-log(0.0035)) ≈ 11.54
F) 25.0 mL HCl contains, 0.025 × 0.258 moles = 0.00645 moles
[OH⁻] remaining = 0.05 × 0.129 - 0.00645 = 0 moles
[tex]Concentration \ of \ [OH^-] = \dfrac{0\, moles}{(0.05 + 0.024) \, L} \approx 0\, M[/tex]
Therefore;
Number of moles of KOH = 0, or the solution is neutralized
[OH⁻] = [H⁺]
Which gives;
- pH = pOH = 7
G) 26.0 mL HCl contains, 0.026 × 0.258 moles = 0.006708 moles
[OH⁻] remaining = 0.05 × 0.129 - 0.006708 = -0.000258 moles
Therefore
Number of moles of H⁺ = 0.000258
[tex]\mathbf{Concentration} \ of \ \mathbf{[H^+] }= \dfrac{0.000258\, moles}{(0.05 + 0.026) \, L} \approx 0.003395\, M[/tex]
- pH of solution = (-log(0.003395)) ≈ 2.469
H) 29.0 mL HCl contains, 0.029 × 0.258 moles = 0.007482 moles
H⁺ remaining = 0.007482 - 0.05 × 0.129 = 0.001032 moles
Therefore
Number of moles of H⁺ = 0.001032
[tex]Concentration \ of \ [H^+] = \mathbf{ \dfrac{0.001032\, moles}{(0.05 + 0.029) \, L}} \approx 0.01306\, M[/tex]
- pH of solution = (-log(0.01306)) ≈ 1.884
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