millimols base = 1,500*0.17 = 255
mmols acid = 1,500*0.32 = 480
add 54 mmols HCl
.........Bz^- + H^+ ==> HBz
I.......255......0.......480
add.............54...............
C.......-54....-54......+54
E.......201.....0.......534
Plug these E line values into the HH equation and solve for pH, then convert to (H^+). I get pH of about 3.8 or so.
A buffer with a pH of 3.93 contains 0.17 M of sodium benzoate and 0.32 M of benzoic acid. What is the concentration of [H ] in the solution after the addition of 0.054 mol of HCl to a final volume of 1.5 L?
1 answer