Question

The next three (3) problems deal with the titration of 301 mL of
0.501 M carbonic acid (H_{2}CO_{3}) (K_{a1}
= 4.3 x 10^{-7}, K_{a2} = 5.6 x 10^{-11})
with 1.5 M KOH.

How many mL of the 1.5 M KOH are needed to raise the pH of the carbonic acid solution to a pH of 6.635?

Answer #1

The balanced reaction

H_{2}CO_{3} + NaOH ? NaHCO_{3} +
H_{2}O

NaHCO_{3} + NaOH ? Na_{2}CO_{3} +
H_{2}O

From the Henderson-Hasselbalch equation:

pH = pKa + log [HCO3-]/[H2CO3]

Ka1 = 4.3 x 10^-7

pKa = - log (4.3 x 10^-7) = 6.366

6.635 = 6.366 + log [HCO3-]/[H2CO3]

[HCO3-]/[H2CO3] = 1.858 ......... Eq1

Given that

[HCO3-] + [H2CO3] = 0.501 M

From Eq 1

1.858 x [H2CO3] + [H2CO3] = 0.501

2.858 x [H2CO3] = 0.501

[H2CO3] = 0.175 M

Again from Eq 1

[HCO3-] = 1.858 x 0.175 = 0.326 M

Moles of HCO3- = molarity x volume

= 0.326 Mol/L x 0.301 L

= 0.098126 mol

From the stoichiometry of the reaction

1 mol HCO3- produced from = 1 mol NaOH

Moles of NaOH = 0.098126 mol

Volume of NaOH = moles/molarity

= 0.098126 mol / 1.5 mol/L

= 0.0654 L x 1000mL/L

= 65.40 mL

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