I think the easy way to do b is to convert everything to millimols and use that to construct an ICE table. I think this way is easier than converting everything to final M.
mmols HBz = mL x M = 250 x 0.125 = 31.25
mmols Bz^ = 250 x 0.147 = 36.75
mmols HCl added = 15.3
.........Bz^- + H^+ ==> HBz
I.....36.75.....0.......31.25
add............15.3............
C....-15.3....-15.3.....+15.3
E.....21.45.....0.......46.55
Substitute into the HH equation. I came up with 3.86
Determine the pH of the following solutions:
a) A 250.0 mL solution that contains 0.125 M benzoic acid and 0.147 M sodium benzoate.
b) The solution in part (a) after 30.0 mL of 0.510 M hydrochloric acid has been added.
a.
Ka = 6.3 x 10^-5
pKa = 4.20
pH = pKa+log(base/acid)
pH = 4.20 + 0.0704
pH = 4.27
b.
HA(aq) + OH-(aq) <----> A-(aq)+H2O(l)
I 0.125M 0.147 - -
C
E
What would i need to do for part b? do i add the hydrochloric acid to the benzoic acid?
1 answer
1 answer