mols NaAc = 8.203/60 = about 0.14 but you need to be more accurate.
M NaAc = 0.14/0.1L = about 1.4M
millimoles NaAc = 20 mL x 1.4 = 28
mmols HAc = 20 mL x 1.OM = 20
mmols HCl = 5 mL x 0.50M = 02.5
..........Ac^- + H^+ ==> HAc
initial...28.....0.......20
add..............2.5.........
change....-2.5..-2.5......+2.5
equil....25.5.....0........2.5
Recalculate the numbers to be more accuratae, substitute those numbers into the Hendeson-Hasselbalch equation and solve for the pH.
If you add 5.0 mL of 0.50 M HCl solution to 20.0 mL to Buffer C, what is the pH of the buffer? (where buffer C is 8.203 g sodium acetate with 100.0 mL of 1.0 M acetic acid)
1 answer
1 answer