With the same initial solution of 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4. You add 16g hydrobromic acid to this solution, what is the new pH?

Question

I tried making this ice chart but it was wrong
16gHBr/80.9119g/mol = 0.1977 mol in 1L
................HCOO^- + H^+ ==> HCOOH
I................0.85.............0................0.75
C..........-0.1977....-0.1977.......+0.1977
E............0.6523....-0.1977......0.5523
I used that and got 3.82 but it was wrong.

With the same initial solution of 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4. You add 60.7g hydrobromic acid to this solution, what is the new pH? Assume a 1L solution.

Answer 1

You didn't substitute correctly. Here is the problem and what you did.
With the same initial solution of 0.75 moles formate and 0.85 moles formic acid to make a buffer solution. The Ka of formic acid is 1.8x10-4. You add 16g hydrobromic acid to this solution, what is the new pH?
I tried making this ice chart but it was wrong
16gHBr/80.9119g/mol = 0.1977 mol in 1L
................HCOO^- + H^+ ==> HCOOH
I................0.85.............0................0.75
C..........-0.1977....-0.1977.......+0.1977
E............0.6523....-0.1977......0.5523
I used that and got 3.82 but it was wrong.
I would have done this.
The problem tells you formic acid is 0.85 and formate (HCOO^-) = 0.75

...............HCOO^- + H^+ ==> HCOOH
I................0.75.............0................0.85
add...........................0.1977........................
C..........-0.1977....-0.1977..........+0.1977
E............0.5523........0................1.0477
Try that. Thanks for showing your work. It showed me EXACTLY what went wrong instantly.

Answer 2

By the way, in the Henderson-Hasslebalch equation, the base is HCOO^- (formate) and the acid is HCOOH (formic acid).

Answer 3

I did that and I got 3.46. But for the second question with 60.7, when I put the numbers in the calculator it says math error.
HCOO- + H^+ ==> HCOOH
I.........0.75..........0...............0.85
Add................0.7502..........
C.....-0.7502...-0.7502.......+0.7502
E....-2 x 10^-4.....0...........1.6005

Answer 4

When you put WHAT into the calculator do you get an math error notation. I'll guess that you're putting ALL of these numbers for the E line into the H-H equation and that's when you get the message. That's because you are putting in a negative number for mols HCOO^- and THAT'S because the 60.7 grams HBr is enough to use up ALL of the formate. Now you don't have a buffer. So you essentially have a solution of HBr from the first part of the problem (0.85 moles) + moles from the formate that made more HBr when you added the HBr. So now the pH is the pH of a solution of straight HBr. Add mols HBr you had at the start + mols HBr formed from adding that 60.7 g HBr and divide by the total volume of the solution. That gives the molarity of HBr and you calculate the pH of that. I'm guessing here that's the problem but you can check it out.