Let's call that HA where C6H50^- is A^- and H^+ is H^+.
...................HA ==> H^+ + A^-
I................0.57.......0..........0
C..............-x...........x..........x
E,,,,,,,,,,,,,0.57-x......x.........x
The problem tells you it is 0.0687% dissociated which means x is 0.57 x 0.000685 = ?
which allows you to calculate x and 0.57-x. You take it from there.
b. Convert (H^+) to pH with pH = -log (H^+)
c. Knowing HBr is a strong acid (it is 100% ionized), (H^+) = (HBr) = ?
In a .57 M solution of HOC6H5, .0684% of the acid has been dissociated.
a) find the concentrations of all aqueous species in the solution at equilibrium
b) find the pH of the solution
c) What concentration of HBr would produce a solution with the same pH as a .57M solution of of HOC6H5?
1 answer
1 answer