In: Chemistry
Acetic acid and its conjugate base acetate can form an acid-base buffer. The pKa of acetic acid is 4.75. How much 10.0 M HNO3 must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH to 4.85 ?
pKa of acetic acid = 4.75
Initial moles of acetic acid = 0.0100 mol/L
Initial moles of sodium acetate = 0.100 mol/L
Let us say we have to add v mol of nitric acid to the buffer solution.
When v of nitric acid is added to the buffer solution, 1 mole of acetate ion reacts with 1 mole nitric acid to give 1 mole of acetic acid.
moles of nitric acid added = v mol
moles of acetic acid = (0.01 + v) mol
moles of sodium acetate =(0.1- v) mol
Hence [sodium acetate] / [acetic acid] = {(0.1- v) / (0.01 + v)}
pH = pKa + log {[sodium acetate] / [acetic acid] }
or, 4.85 = 4.75 + log {(0.1- v) / (0.01 + v)}
or, 0.1 = log {(0.1- v) / (0.01 + v)}
or, (0.1- v) / (0.01 + v) = 100.1 = 1.259
or, 0.1- v = 1.259 (0.01 + v) = 0.01259 + 1.259v
or, 1.259v +v = 0.1- 0.01259 = 0.08741
or, v= 0.03869 mol
Now 0.03869 mol of 10 M nitric acid = 0.03869 mol x ( 1 L /10 mol) = 0.003869 L = 3.869 ml
We need to add 3.869 ml of 10 M nitric acid.
or, 0.00259 v =
0.1 - X = 1.259(.01 + X) --> 2.259X = 0.1 - 0.01259