Question

Based on the Ka, what are the amounts (mL) and identies of 0.20 M carbonate solutions and water must be mixed together to make 300 mL of a buffer with a pH of 10.50 and a concentration if 0.10M?

H2CO3 <————> H+ + HCO3- Ka1 = 4.7 x 10^-7

HCO3- <————> H+ CO3^2 Ka2= 4.7 x 10^-11

Answer 1

The initial concentration of carbonate solution = 0.2 M

The amount of buffer solution = 300 mL

The pH of buffer solution = 10.5

The concentration of buffer solution = 0.1 M

The no. of mmol of buffer solution = 0.1 mmol/mL * 300 mL = 30 mmol

i.e. n_HCO3^- + n_CO3^2- = 30 .......... Equation 1

Where n = corresponding no. of mmol

According to Henderson-Hasselbalch equation, the following expression can be written.

pH = pKa2 + Log(n_CO3^2-/n_HCO3^-)

i.e. 10.5 = -Log(4.7 x 10^-11) + Log(n_CO3^2-/n_HCO3^-)

i.e. 10.5 = 10.3 + Log(n_CO3^2-/n_HCO3^-)

i.e. Log(n_CO3^2-/n_HCO3^-) = 0.2

i.e. n_CO3^2-/n_HCO3^- = 1.5 ......... Equation 2

From Equations 1 and 2, you can express as shown below.

(30 - n_HCO3^-)/n_HCO3^- = 1.5

i.e. n_HCO3^- = 12 and nCO₃^2- = 18

Formula: M1V1 (dilute solution) = M2V2 (concentrated solution)

i.e. 0.1 M * 300 mL = 0.2 M * V2

i.e. V2 = 150 mL, the volume of 0.2 M carbonate solutions

Now, the volume of water that must be added = 300 - 150 = 150 mL