A solution contains 0.0273 M HNO3, 0.0302 M HI, and 0.320 M formic acid, HCOOH. What is the pH?
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!
UPDATE: I got the last one right, but I still can't get the first one.
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.
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.
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.
To determine the pH of a solution, you need to consider the strengths of the acids and bases present and their respective concentrations. The pH of a solution can be calculated using the equation:
pH = -log10[H+]
To find [H+], we need to consider the dissociation of the acids and bases present in the solutions.
1. For the first question, the equation for the dissociation of HNO3 is:
HNO3 -> H+ + NO3-
Since HNO3 is a strong acid, it dissociates completely in solution. Therefore, the concentration of [H+] is equal to the concentration of HNO3 in this case.
[H+] = 0.0273 M
Similarly, the equation for the dissociation of HI is:
HI -> H+ + I-
Again, since HI is a strong acid, its concentration serves as the concentration of [H+] in the solution.
[H+] = 0.0302 M
To determine the pH of the solution, we'll need to calculate the total concentration of [H+]. Since the concentrations of HNO3 and HI are in the same solution, we can simply add them together.
Total [H+] = [H+] from HNO3 + [H+] from HI
= 0.0273 M + 0.0302 M
= 0.0575 M
Now, we can calculate the pH using the equation:
pH = -log10[H+]
= -log10(0.0575)
≈ 1.2406
Therefore, the pH of the solution is approximately 1.2406.
2. For the second question, we need to consider the dissociation of the bases present.
The dissociation equation of Ba(OH)2 is:
Ba(OH)2 -> Ba2+ + 2OH-
Since Ba(OH)2 is a strong base, it fully dissociates in solution. Therefore, the concentration of [OH-] is 2 times the concentration of Ba(OH)2.
[OH-] = 2 * 0.160 M
= 0.320 M
Next, we need to consider the dissociation of ammonia (NH3):
NH3 + H2O -> NH4+ + OH-
Ammonia is a weak base, so we need to consider its equilibrium constant, Kb. The Kb expression for ammonia is:
Kb = [NH4+][OH-] / [NH3]
However, we'll assume complete dissociation of NH3, so the concentration of NH3 will be the same as [OH-].
[OH-] = 0.360 M
Now, we can calculate the concentration of [H+] using the equation:
Kw = [H+][OH-] = 1.0 x 10^-14
Since [H+] = [OH-] in a neutral solution, we can solve for [H+] using the equation:
[H+] = √(Kw/[OH-])
= √(1.0 x 10^-14 / 0.360)
≈ 5.165 x 10^-8 M
Finally, we can calculate the pH using the equation:
pH = -log10[H+]
= -log10(5.165 x 10^-8)
≈ 7.286
Therefore, the pH of the solution is approximately 7.286.