Question

1.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 8g sodium hydroxide to this solution, what is the new pH?
2. 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?
3. 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

HCOOH is formic acid = 0.85 moles in 1 L = 0.85 M
HCOO^- is formate ion. = 0.75 moles in 1 L = 0.75 M
mols NaOH = grams/molar mass = 8/40 = 0.20 mol in 1 L = 0.20 M
The equation is
...................HCOOH + OH^- ==> HCOO^- + H2O
I....................0.85...........0...............0.75............0
add...............................0.20........................................
C..................-0.20........-0.20............+0.20.................
E..................0.65..............0..............0.95.....................
Plug the E line into the pH = pKa + log [(base)/(acid)] and solve for pH.

The others follow the same procedure except that when you're adding and acid to the buffer the equation is HCOO^- + H^+ ==> HCOOH and the ICE chart changes. Post your work if you get stuck.

Answer 2

Could you help with the second 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.