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.

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.

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.

thank you!

a) To calculate the pH of the buffer solution, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa of the weak acid and the ratio of the weak acid to its conjugate base.

The Henderson-Hasselbalch equation is given by:
pH = pKa + log([conjugate base]/[acid])

In this case, formic acid (HCOOH) is the weak acid, and sodium formate (HCOONa) is its conjugate base. The pKa of formic acid is given as 1.8 x 10^-4.

We are given that the buffer solution contains 0.4 mol of formic acid (HCOOH) and 0.6 mol of sodium formate (HCOONa) in 1 L of solution.

First, we need to calculate the concentrations of the acid and the conjugate base. Since we have 0.4 mol of formic acid and 0.6 mol of sodium formate in 1 L of solution, the concentrations are:
[acid] = 0.4 M
[conjugate base] = 0.6 M

Now we can substitute these values into the Henderson-Hasselbalch equation:
pH = 1.8 x 10^-4 + log(0.6/0.4)
pH = 1.8 x 10^-4 + log(1.5)

Using logarithm properties, log(1.5) can be calculated as approximately 0.1761.

Thus, the pH of the buffer solution is:
pH = 1.8 x 10^-4 + 0.1761
pH ≈ 2.1761

b) When 100 mL of the buffer solution is diluted to 1 L with pure water, the pH remains constant because the ratio of the weak acid to its conjugate base remains the same. The Henderson-Hasselbalch equation shows that the pH of a buffer is dependent on this ratio. Dilution with water does not affect the ratio of the acid to the conjugate base, so the pH remains constant.

c) To determine the pH of the resulting solution after adding 5 mL of 1 M HCl to 100 mL of the original buffer solution, we need to consider the reaction between HCl and the weak acid (formic acid) in the buffer.

HCl is a strong acid and completely ionizes in water, producing H3O+ ions. When HCl reacts with formic acid, it forms formate ions (HCOO-) and additional H3O+ ions.

To calculate the resulting [H3O+] of the solution, we need to consider the initial concentrations of H3O+ from HCl and the H3O+ produced by the reaction with formic acid.

First, let's calculate the concentration of H3O+ from HCl:
[H3O+] from HCl = 5 mL x 1 M / (100 mL + 5 mL)
[H3O+] from HCl = 0.05 M

Now, let's consider the reaction between HCl and formic acid:
HCl + HCOOH → HCOO- + H3O+

From the balanced reaction equation, we can see that 1 mol of HCl reacts with 1 mol of formic acid to produce 1 mol of formate ion (HCOO-) and 1 mol of H3O+.

Since the original buffer solution contains 0.4 mol of formic acid, the resulting concentration of formate ions and additional H3O+ ions will be 0.4 M.

Therefore, the total [H3O+] in the resulting solution is the sum of the [H3O+] from HCl and the additional [H3O+] from formic acid:
[H3O+] = [H3O+] from HCl + [H3O+] from formic acid
[H3O+] = 0.05 M + 0.4 M
[H3O+] = 0.45 M

Thus, the [H3O+] of the resulting solution is 0.45 M.

d) To determine the [H3O+] of the resulting solution when an 800 mL sample of 2 M formic acid is mixed with 200 mL of 4.8 M NaOH, we need to consider the neutralization reaction between the acid and the base.

The balanced equation for the reaction between formic acid (HCOOH) and sodium hydroxide (NaOH) is:
HCOOH + NaOH → HCOONa + H2O

From the balanced reaction equation, we can see that 1 mol of formic acid reacts with 1 mol of sodium hydroxide to produce 1 mol of sodium formate (HCOONa) and 1 mol of water. The reaction is 1:1.

Using the given concentrations and volumes, we can calculate the number of moles of formic acid and sodium hydroxide used in the reaction:

Moles of formic acid = 800 mL × (2 mol/L) = 1.6 mol
Moles of sodium hydroxide = 200 mL × (4.8 mol/L) = 0.96 mol

Since the moles of formic acid and sodium hydroxide are in a 1:1 ratio, the limiting reagent is sodium hydroxide.

Therefore, the moles of sodium hydroxide fully react with the moles of formic acid, leaving no excess.

The resulting solution will contain 1.6 - 0.96 = 0.64 mol of formic acid that did not react, and the volume of the resulting solution is 800 mL + 200 mL = 1000 mL = 1 L.

Now, we can calculate the concentration of formic acid in the resulting solution:
[formic acid] = 0.64 mol / 1 L = 0.64 M

Given that formic acid is a weak acid with a Ka value of 1.8 x 10^-4, we can use the Henderson-Hasselbalch equation to calculate the pH of the resulting solution, similar to part a).

pH = pKa + log([conjugate base]/[acid])

In this case, formic acid (HCOOH) is still the weak acid, and its conjugate base is sodium formate (HCOONa). The pKa value remains 1.8 x 10^-4, and the concentration of formic acid is 0.64 M.

Substituting these values into the Henderson-Hasselbalch equation:
pH = 1.8 x 10^-4 + log(0.6/0.4)
pH = 1.8 x 10^-4 + log(1.5)

Using logarithm properties as previously mentioned, log(1.5) ≈ 0.1761.

Thus, the pH of the resulting solution is:
pH = 1.8 x 10^-4 + 0.1761
pH ≈ 2.1761