Based on the Ka, what are the amounts (mL) and identies of 0.20 M carbonate solu

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Explanation / Answer

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. nHCO3- + nCO32- = 30 .......... Equation 1

Where n = corresponding no. of mmol

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

pH = pKa2 + Log(nCO32-/nHCO3-)

i.e. 10.5 = -Log(4.7 x 10-11) + Log(nCO32-/nHCO3-)

i.e. 10.5 = 10.3 + Log(nCO32-/nHCO3-)

i.e. Log(nCO32-/nHCO3-) = 0.2

i.e. nCO32-/nHCO3- = 1.5 ......... Equation 2

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

(30 - nHCO3-)/nHCO3- = 1.5

i.e. nHCO3- = 12 and nCO32- = 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

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