Can anyone help me with this problem? "Find the pH of the two equivalence points and the volume (mL) of 0.0378 M KOH needed to reach them in the titration of 17.3 mL of 0.130 M H2CO3. 1st Equivalent Point ___ mL KOH ___ pH 2nd Equivalent Point ___ mL KOH ___ pH

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