How would I find the pH of these? I found it for d and e, but I don't know how to do the rest
a) 0.25M carbonic acid
b) 0.075M CN-
c) 0.15M F-
d) 0.015M HCl (I got 1.8)
e) 0.015 M NaOH (I got 12)
f) 0.050 M acetic acid
For e I think you threw away one of the digits. I would have kept it at 1.82 (when taking logs the 1 is there as a decimal placer; therefore, you are allowed two significant figures and 1.82 would be ok. Same thing for e. That would be 14-1.82 = ?
a. and f are done the same way. For f,
........HAc --> H^+ + Ac^-
I......0.05.....0......0
C.......-x......x......x
E....0.05-x.....x......x
Ka = (H+)(Ac^-)/(HAc)
Substitute the E line into Ka expression, solve for x = H^+ and convert to pH.
a is done exactly the same way BUT you have both ka1 and ka2. In most cases the k values are so far apart that you use k1 ONLY and solve it the same way as f above. The reason you can do this is because the contribution from k2 is so small it is negligible.
b and c are done te same way. They are bases and they hydrolyze in water as follows.
..........CN^- + HOH ==> HCN + OH^-
I.......0.075.............0.....0
C.........-x..............x.....x
E......0.075-x............x.....x
Kb for CN^- = (Kw/Ka for HCN) = (x)(x)/(0.075-x)
Solve for x = (OH^-) and convert to pH.