How would I find the pH of 0.25 M carbonic acid? also the pH for 0.075 M CN^-?

Answer 1

H2CO3 has two ionization constants; i.e., ka1 and ka2. Generally the Ka values for diprotic acids is far apart and you simply treat these as two separate ionizations; the second one contributes little so you ignore it.
..........H2CO3 ==> H^+ + HCO3^-
I........0.23.......0......0
C..........-x.......x......x
E........0.23-x.....x......x

Substitute the E line into Ka1 and solve for x = (H^+), then convert to pH.

For CN^-, the pH of the solution is determined by the hydrolysis of the "base".
...........CN^- + HOH ==> HCN + OH^-
I.........0.075...........0......0
C..........-x.............x......x
E........0.075-x..........x......x

Then Kb for CN = (Kw/Ka for HCN) = (x)(x)/(0.075-x) and solve for x = (OH-) and convert that to pH.

Answer 2

thanks!