Question

You need 100.0 mL of a buffer at pH 10.50 based on carbonate (H2CO3: Ka1= 4.3 x 10-7, Ka2= 5.6 x 10-11). You have available the following materials: solid sodium bicarbonate (84.0066g/mol), 2.000 M HCl, and 2.000 M NaOHand of course, deionized water. To ensure adequate buffering capacity, the buffer component present at lower concentration should be at 0.200 M. Provide a recipe indicating volumes of liquids and/or masses of solids for the technician to use in making this buffer.

Answer: 18 mL + 4.7 g NaHCO3 + balance deionzed water to 100.0 mL

-Please explain, Thank you.

Answer 1

Volume of buffer = 100mL

pH = 10.50

Ka1 = 4.3 x 10-7, pKa1 = 6.36

Ka2= 5.6 x 10-11 pKa2 = 10.25

As per Pka (near to desried pH value) values we will take sodium carbonate and sodium bicarbonate solution for making buffer

Now as we are provided with solid sodium bicarbonate and solution of HCl and NaOH

so we will use NaOH and sodium bicarbonate for preparation of buffer

NaOH + NaHCO3 --> Na2CO3 + H2O

Desired concentration is not given just volume is given

We will use Henersen equaiton which is used to caluculate pH of any buffer

pH = pKa + log[Salt] / [Acid]

10.5 = 10.25 + log[Na2CO3 / NaHCO3]

0.25 = log[Na2CO3 / NaHCO3]

Taking antilog

[Na2CO3 / NaHCO3] = 1.778

So the ratio of concentration should be =1.778

The buffer component at lower concentration is NaHCO3

Its concentration = 0.2

So concentration of Na2CO3 = 1.778 X 0.2 = 0.355 M

Moles = 0.355 X Volume = 0.0355

We are provided with NaOH of concentration 2M

as per stoiciometry of above equation

1moles of NaOH will react with 1 moles of NaHCO3 to give 1 moles of Na2CO3

so for 0.0355 moles we need 0.0355 moles of NaOH + 0.0355 moles of Na2CO3

As per concentration of given NaOH , for 2 moles of NaOH we will need 1Litres of solution

so for 0.0355 moles we will take = 0.0355/ 2 = 0.0177Litres or 18 mL

Now let us calculate the total amount of sodium bicarbonate required

Molarity of NaHCO3 = 0.2 M

So moles = 0.02 moles

In addiiton we need 0.0355 moles for preparation of Na2CO3 by reaction with NaOH
(as calculated above)

So total moles = 0.0355 X 0.02 = 0.055

Moles = Mass / Molecular weight =

Mass = Moles X mol wt = 0.055 X 84.0066 = 4.662grams = 4.7 grams

and will make up the solution to 100mL