Question

What are the amounts (mL) and identies of 0.30 M carbonate solutions and water must be mixed together to make 250 mL of a buffer with a pH of 10.50 and a concentration if 0.10M?

HCO3- <————> H+ CO3^2 Ka1= 4.7 x 10^-11

H2CO3 <————> H+ + HCO3- Ka2 = 4.7 x 10^-7

Answer 1

We have to make use of the Ka1 reaction because the pH is close to this pKa value.

For reaction 1:

pKa1 = -log(Ka1) = 11-log(4.7) = 10.33

Using Henderson Hasselbach equation:

pH = pKa + log(moles of CO₃^2-/moles of HCO₃^-)

Since the conc of all the carbonate solutions is the same and equal to 0.30 M, so the number of moles is directly proportional to the volume of the solution used.

So the above equation can be written as:

pH = pKa + log(volume of CO₃^2-/volume of HCO₃^-)

Putting values:

10.50 = 10.33 + log(volume of CO₃^2-/volume of HCO₃^-)

Solving we get:

(volume of CO₃^2-/volume of HCO₃^-) = 1.479

Also,

Volume of CO₃^2- + Volume of HCO₃^- = 250 mL

Solving these two equations we get:

Volume of 0.30 M CO₃^2- solution needed = 149.15 mL

Volume of 0.30 M HCO₃^- solution needed = 100.85 mL

Hope this helps !