For a solution that is 0.20 M HCN(aq) with a Ka=4.9x10^-10

Question

a. calculate the [H+]
b. calculate the [OH-]
c. what's the pH?

Please show your work on how to solve it. Thank you

Answer 1

.............HCN ==> H^+ + CN^-
initial....0.20M.....0.....0
change......-x.......+x.....+x
final.... 0.20-x......x.....x

Ka = ((H^+)(CN^-)/(HCN)
Substitute from the above and solve for (H^+).
pH = -log(H^+)
pH + pOH = pKw = 14
Post your work if you get stuck.

Answer 2

Thanks! That was helpful.

Answer 3

thunder blew a 3-1 lead