Asked by Parveen
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.
You should specify the amount of sodium carbonate and sodium hydrogen carbonate that you would use.
You should specify the amount of sodium carbonate and sodium hydrogen carbonate that you would use.
Answers
Answered by
DrBob222
Use the Henderson-Hasselbalch equation.
pH = pk2 + log(base/acid)
10.80 = 10.33 + log B/A
B/A = 2.95
CO3^2- = 2.95*HCO3^-
Now you need to pick a number for HCO3^- that will dissolve in 100 mL H2O (say 1.00 g, convert to mols, and calculate g CO3^2- needed. Dissolve the two salts in 100 mL H2O. Plug those numbers back into the HH equation to make sure the pH is 10.80.
Also, look up the solubilities of NaHCO3 and Na2CO3 to make sure they will dissolve in 100 mL H2O.
pH = pk2 + log(base/acid)
10.80 = 10.33 + log B/A
B/A = 2.95
CO3^2- = 2.95*HCO3^-
Now you need to pick a number for HCO3^- that will dissolve in 100 mL H2O (say 1.00 g, convert to mols, and calculate g CO3^2- needed. Dissolve the two salts in 100 mL H2O. Plug those numbers back into the HH equation to make sure the pH is 10.80.
Also, look up the solubilities of NaHCO3 and Na2CO3 to make sure they will dissolve in 100 mL H2O.
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