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Answer:
1st Equivalent Point
59.5 mL KOH 8.34 pH
2nd Equivalent Point
119 mL KOH 11.25 pH
Explanation:
These are the two neutralization reactions:
H₂CO₃ + OH⁻ ⇄ HCO₃⁻ + H₂O
HCO₃⁻ + OH⁻ ⇄ CO₃⁻² + H₂O
For the first equivalence point we need to know that
mili mol of base = mili mol of acid
0.0378 M . volume (mL) = 17.3 mL . 0.130M
Volume (mL) = (17.3 mL . 0.130M) / 0.0378M = 59.5 mL
Global reaction is:
H₂CO₃ + 2OH⁻ ⇄ CO₃⁻² + 2H₂O
0.0378 M . volume (mL) = 2 . 17.3mL . 0.130M
Volume(mL) = 119 mL
The first equivalence point has only the HCO₃⁻ as predominant, so the pH will be the media, of the two pKa from the H₂CO₃
(pKa₁ + pKa₂) / 2 = (6.35 + 10.32) / 2 = 8.34 → pH₁
In the second equivalence point the predominant is the CO₃⁻² which is a base. First of all, we determine the [CO₃⁻²]
[CO₃⁻²] = mili moles of acid / initial volume + volume for the 2nd point
[CO₃⁻²] = 0.130 M . 17.3 mL / 17.3 mL + 119mL = 0.0165 M
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb₁
pKb₁ = 14 - pKa₂ → 14 - 10.32 = 3.68
Kb₁ = 10⁻³°⁶⁸ = 2.09 ₓ10⁻⁴
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb₁
0.0165 - x x x
Kb₁ = x² / 0.0165 - x
2.09 ₓ10⁻⁴ ( 0.0165 - x ) - x² = 0
This is a quadractic. We can not use, only 0.0165 in the denominator because the Kb is in order of 10⁻⁴)
3.4485×10⁻⁶ - 2.09×10⁻⁴x - x²
a= -1
b= - 2.09×10⁻⁴
c = 3.4485×10⁻⁶
(- b +- √(b² - 4ac) ) / 2
x = 1.75×10⁻³
pOH = - log [OH⁻] =2.75
pH = 14 - pOH → 11.25
The first Equivalent Point of 59.5 mL KOH 8.34 pH. The second Equivalent Point of 119 mL KOH 11.25 pH.
Neutralization reactions:
These are the two neutralization reactions:
H₂CO₃ + OH⁻ ⇄ HCO₃⁻ + H₂O
HCO₃⁻ + OH⁻ ⇄ CO₃⁻² + H₂O
GIVEN:
0.0378 M . volume (mL) = 17.3 mL . 0.130M
Volume (mL) = (17.3 mL . 0.130M) / 0.0378M = 59.5 mL
Overall reaction is:
H₂CO₃ + 2OH⁻ ⇄ CO₃⁻² + 2H₂O
0.0378 M . volume (mL) = 2 . 17.3mL . 0.130M
Volume(mL) = 119 mL
The first equivalence point has only the HCO₃⁻ as predominant, so the pH will be the media, of the two pKa from the H₂CO₃
(pKa₁ + pKa₂) / 2 = (6.35 + 10.32) / 2 = 8.34 → pH₁
In the second equivalence point the predominant is the CO₃⁻² which is a base.
First of all, we determine the [CO₃⁻²]
[CO₃⁻²] = mili moles of acid / initial volume + volume for the 2nd point
[CO₃⁻²] = 0.130 M . 17.3 mL / 17.3 mL + 119mL = 0.0165 M
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb₁
pKb₁ = 14 - pKa₂ → 14 - 10.32 = 3.68
Kb₁ = 10⁻³°⁶⁸ = 2.09 ₓ10⁻⁴
CO₃⁻² + H₂O ⇄ HCO₃⁻ + OH⁻ Kb₁
0.0165 - x x x
Kb₁ = x² / 0.0165 - x
2.09 ₓ10⁻⁴ ( 0.0165 - x ) - x² = 0
3.4485×10⁻⁶ - 2.09×10⁻⁴x - x²
a= -1
b= - 2.09×10⁻⁴
c = 3.4485×10⁻⁶
Using the formula for quadratic equation:
(- b +- √(b² - 4ac) ) / 2
x = 1.75×10⁻³
pOH = - log [OH⁻] =2.75
pH = 14 - pOH →
pH= 11.25
Find more information about Neutralization reaction here:
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