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.
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.
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.
By the way, in the Henderson-Hasslebalch equation, the base is HCOO^- (formate) and the acid is HCOOH (formic acid).
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
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.
To calculate the new pH of the solution, we need to consider the effect of adding hydrobromic acid (HBr) to the buffer solution.
First, let's determine the number of moles of HBr added:
60.7g HBr / molar mass of HBr (80.9119 g/mol) = 0.750 mol HBr
Since HBr is a strong acid, it will completely dissociate in water to produce H+ ions. Therefore, the concentration of H+ ions after adding HBr is 0.750 mol/L.
Next, we need to calculate the change in concentration of formate ion (HCOO-) and formic acid (HCOOH). Since 1 mol of HBr reacts with 1 mol of formate ion, the change in concentration of HCOO- is -0.750 mol/L. Similarly, since 1 mol of HBr reacts with 1 mol of formic acid, the change in concentration of HCOOH is also -0.750 mol/L.
To find the final concentrations of HCOO- and HCOOH, we subtract the changes from their initial concentrations:
[HCOO-] = 0.75 mol/L - 0.750 mol/L = 0 mol/L
[HCOOH] = 0.85 mol/L - 0.750 mol/L = 0.1 mol/L
Remember that a buffer solution consists of a weak acid and its conjugate base. In this case, formic acid (HCOOH) is the weak acid and formate ion (HCOO-) is its conjugate base.
Now, using the Henderson-Hasselbalch equation, we can calculate the new pH of the buffer solution:
pH = pKa + log([A-]/[HA])
Given that the Ka of formic acid (HCOOH) is 1.8x10^-4, we can calculate the pKa, which is the negative logarithm of the Ka:
pKa = -log(1.8x10^-4) = 3.74
Substituting the values into the equation, we have:
pH = 3.74 + log(0/0.1)
Since the concentration of formate ion is now 0 mol/L, the logarithm term becomes undefined. This is because the logarithm of 0 is not defined.
Therefore, the new pH of the solution cannot be calculated in this case.