Question

How many liters of 0.505 M NaOH will be needed to raise the pH of 0.383 L of 5.99 M carbonic acid (H₂CO₃) to a pH of 9.983?

Answer 1

pH of 5.99 M carbonic acid (H₂CO

H₂CO₃ H⁺ + HCO₃^-

Ka₁ = [H⁺] [HCO₃^-] / [H₂CO₃]
​or, 4.3 x 10^-7 = (x)(x) / (5.99 - x)
or, 4.3 x 10^-7 = x² / 5.99 (assuming x << 0.383)
or, x² = (4.3 x 10^-7) x 5.99
​or, x² = 2.58 x 10^-6
​or, x = 1.6 x 10^-3

[H⁺] = x = 1.6 x 10^-3

^​pH = - log [H⁺]
​ = - log (1.6 x 10^-3)
​ = 2.796

For, carbonic acid Ka2 = 4.7 x 10^-11

pKa2 = - log (4.7 x 10^-11) = 10.33

pH < 10.33, so discarding Ka2 factor.

Change in pH = 9.983 - 2.796 = 7.187

- log [H+] = pH
or, - log [H+] = 7.187
or, [H+] = 10^-7.187
or, [H+] = 6.5 x 10^-8 M

Volume = 0.383 L

So, moles of [H+] = (6.5 x 10^-8 M) x 0.383 L) = 2.49 x 10^-8 moles

So, 2.49 x 10^-8 moles of [H+] will be neutralize with 2.49 x 10^-8 moles​ of [OH-]

Molarity of [OH-] = 0.505 M

So, volume of [OH-] = (2.49 x 10^-8 moles) / 0.505 M = 4.93 x 10^-8 L

s = 1.0 x 10-4 = [H+] (assumption that s << 0.025 not too bad)

pH = - log [H+] = - log (1.0 x 10^-4) = 4.0
% dissociation = (1.0 x 10^-4 / 0.025 ) x 100 = 0.4 %

The second proton that comes off HCO31- can be neglected because the Ka2 is 5.6 x 10^-11 and doesn't contribute enough H+ to affect the pH significantly.

2) Equation:
2NaOH + H2CO3 → Na2CO3 + 2H2O
2mol NaOH reacts with 1 mol H2CO3