A solution contains 0.0273 M HNO3, 0.0302 M HI, and 0.320 M formic acid, HCOOH. What is the pH?

Question

A solution contains 0.160 M Ba(OH)2 (strong base) and 0.360 M ammonia, NH3 (weak base). What is the pH?

I'm really not sure how to do this. I tried and got .443 and 12.91 respectively, but they were both wrong. If you could help, it would be much appreciated. Thanks!

Answer 1

UPDATE: I got the last one right, but I still can't get the first one.

Answer 2

I would add the HNO3 and HI to find H^+ from the two strong acids. Use that as a common ion and Ka for HCOOH to determine H^+ from the HCOOH. Add all to determine total H^+ and convert to pH. If you still don't get it show what you did and show the Ka you use for HCOOH.

Answer 3

I still don't get it. I used an ICE table to get x=sqrt(.32*1.8e-4)=.0076
Then I added .0076+.0273+.0302 to get.0651 and then I got the -log and got 2.73, which wasn't right. I used 1.8e-4 as ka.

Answer 4

Note that sentence in my instructions where I said use the total H^+ from the strong acids "as a common ion." You didn't do that. The H^+ from HCOOH is x BUT the total H^+ is x + 0.0575. The whole idea here is that the H^+ from the strong acids suppresses the ionization of HCOOH so the contribution from HCOOH is ONLY 0.001 (you would get 0.0076 otherwise).
(HNO3) = 0.0273M
(HI) = 0.0302M
total H^+ = 0.0273+0.0302 = 0.0575M

..........HCOOH ==> H^+ + HCOO^-
I.........0.320....0.0575......0
C..........-x.......x.........x
E........0.320-x...0.0575+x....x

1.8E-4 = (H^+)(HCOO^-)/(HCOOH)
1.8E-4 = (0.0575)(x)/(0.320-x)
If I make the simplifying assumptions I come out with x = 0.001 so total H^+ = 0.001 + 0.0273+0.0302 = 0.0585 M and covert to pH = 1.23.
If I solve the quadratic I come out with
x = 0.00098 (almost the same) as with the assumptions) and total H^+ = 0.05848; pH = 1.23.