Plan how you would make 100mL of a buffer solution with a pH of 10.80 to be made using only sodium carbonate, sodium hydrogen carbonate and water.

Question

You should specify the amount of sodium carbonate and sodium hydrogen carbonate that you would use.

Below is what i got up to then im not sure how to find the amount of each

pH = pk2 + log(base/acid)
10.80 = 10.33 + log B/A
B/A = 2.95
CO3^2- = 2.95*HCO3^-

thx

Answer 1

I didn't check any of your calculations. There is no indication of what concentration they want the buffer; therefore, I would assume some convenient (and reasonable) amount of one of the salts, say Na2CO3 = 0.05 mols (which would make the Na2CO3 about 0.5M).
That will be 0.05 x 106 = about 5 g Na2CO3 and 0.05/2.95 = about 0.02 mols (0.2M) NaHCO3 or about 0.02 x 84 = about 1.5 g NaHCO3. First you need to calculate these with a little better accuracy, then plug each into the HH equation and see if you end up with a pH of 10.80.
There is another way to do it; i.e., assume that the concentration of the buffer is to be 0.1M which give you a second equation.
eqn 1 is base = 2.95*acid
acid + base = 0.1
Solve the two equation simultaneously to get acid and base, then convert those to grams. You won't get the same answer from these two calculations because the first one works out to be about 0.7M and the last one we start with 0.1M. I would recommend the latter because 0.7M sounds high to me for making a buffer solution.