a) Use the Henderson-Hasselbalch equation shown on the next line below.
b) pH = pKa + log[(base)/(acid)]
When the solution is diluted you change the base concn and you change the acid concn but does the ratio change?
c) Addition of HCl increases concn acid and decreases formate (base).
d) Addition of NaOH, decreases formic acid and increases formate (base).
Post your work if you get stuck.
A buffer solution contains .4mol of formic acid, HCOOH and a .6mol of sodium formate, HCOONa, in 1L of solution. Ka of formic acid is 1.8 x 10^-4.
a) calculate pH
b) if 100ml of this buffer solution is diluted to a volume of 1L with pure water, the pH does not change. Discuss why the pH remains constant on dilution.
c) a 5ml sample of 1M HCl is added to 100ml of the original buffer solution. Calculate [H3O+] of the resulting solution.
d) a 800ml sample of 2M formic acid is mixed with 200ml of 4.8M NaOH. Calculate the [H3O+] of the resulting solution.
4 answers
i understand a and b, but how do i calculate parts c and d?
c) The way a buffer works:
1. If strong acid is added (HCl for example), the base (HCOO^-) takes over and ties up the H^+ by forming HCOOH. So the HCOOH concn is increased and the HCOO^- is decreased by the same amount, the log base/acid term changes and pH changes slightly but not too much because instead of adding a strong acid the system thinks we just added a little more of the weak formic acid.
How much HCOO^- (formate) did we have?
We had 100 mL x 0.6 M = 60 mmoles formate.
We had 100 mL x 0.4 M = 40 mmoles HCOOH.
(You don't need to convert to moles, you CAN still use molarities).
So we add 5 mL x 1 M HCl = 5 mmoles HCl.
That means the new mmoles HCOOH is 40+5 = 45. The new mmoles formate (HCOO^-) is 60-5= 55.
Plug those values into the HH equation and solve for pH.
2. When NaOH is added to a formic acid/formate buffer, it works this way.
The NaOH is neutralized by the formic acid to form more sodium formate. So the acid is decreased, the formate is increased. This is done the same way as part c.
1. If strong acid is added (HCl for example), the base (HCOO^-) takes over and ties up the H^+ by forming HCOOH. So the HCOOH concn is increased and the HCOO^- is decreased by the same amount, the log base/acid term changes and pH changes slightly but not too much because instead of adding a strong acid the system thinks we just added a little more of the weak formic acid.
How much HCOO^- (formate) did we have?
We had 100 mL x 0.6 M = 60 mmoles formate.
We had 100 mL x 0.4 M = 40 mmoles HCOOH.
(You don't need to convert to moles, you CAN still use molarities).
So we add 5 mL x 1 M HCl = 5 mmoles HCl.
That means the new mmoles HCOOH is 40+5 = 45. The new mmoles formate (HCOO^-) is 60-5= 55.
Plug those values into the HH equation and solve for pH.
2. When NaOH is added to a formic acid/formate buffer, it works this way.
The NaOH is neutralized by the formic acid to form more sodium formate. So the acid is decreased, the formate is increased. This is done the same way as part c.
thank you!
1 answer