What are the amounts (mL) and identies of 0.30 M carbonate solutions and water m

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 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
HCO3- <————> H+ CO3^2 Ka1= 4.7 x 10^-11
H2CO3 <————> H+ + HCO3- Ka2 = 4.7 x 10^-7

Explanation / Answer

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 CO32-/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 CO32-/volume of HCO3-)

Putting values:

10.50 = 10.33 + log(volume of CO32-/volume of HCO3-)

Solving we get:

(volume of CO32-/volume of HCO3-) = 1.479

Also,

Volume of CO32- + Volume of HCO3- = 250 mL

Solving these two equations we get:

Volume of 0.30 M CO32- solution needed = 149.15 mL

Volume of 0.30 M HCO3- solution needed = 100.85 mL

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