Question

Consider a the titration of 0.905 L of 0.679 M carbonic acid (H₂CO₃) with 1.65 M NaOH. What is the pH at the second equivalence point of the titration?

Answer 1

moels of H2CO3 = 0.905 x 0.679 = 0.6145

moles of NaOH = 2 x 0.6145 = 1.229

moles = moalrity x volume

1.229 = 1.65 x volume

volume of NaOH = 0.745 L

concentration of CO32- = 0.905 x 0.679 / 0.905 + 0.745

                                       = 0.3724 M

CO32- +   H2O    -----------> HCO3-   +   OH-

0.3724                                        0              0

0.3724- x                                   x                x

Kb = x^2 / 0.3724 - x

2.128 x 10^-4 = x^2 / 0.3724 - x

x = 8.796 x 10^-3

[OH-] = 8.796 x 10^-3 M

pOH = -log (8.796 x 10^-3 )

pOH = 2.06

pH = 11.94