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.

Question

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.