The pK3 of formic acid is 3.75

A)what is the pH of a buffer in which formic acid and sodium formate have equimolar concentration?
B)what is the pH of a solution in which the sodium formate is 10M and the formic acid is 1M?

a. pH = pKa + log(base)/(acid)

pH = 3.75 + log(x)/(x) (the problem just says base = acid so I've called that x.
pH = 3.75 + log(1)
pH = 3.75 + 0
pH = 3.75

b.
HCOOH = formic acid = acid = 1.0M
HCOONa = sodium formate = base = 10M
pH = 3.75 + log (10)/(1)
Solve for pH You should get 4.75.
Acids/bases are described with the Bronsted-Lowry theory as
HA = acid
HA + H2O ==> H3O^+ + A^-
HA is the acid and A^- is its conjugate base.
H2O is the base and H3O^+ is its conjugate acid.
How do you know which is which? The acid is the one with more atoms; the base is the one with fewer H atoms. Look at HA. HA and A. HA is the acid because it has more H atoms than the A. A is the base becasuse it has fewer H atoms than HA.
H2O/H3O^+. H3O^+ has more H atoms; it's the acid. H2O has fewer H atoms; it's the base.

I don't believe formic acid has a pK3. You must mean pKa.

a. pH = pKa + log (base)/(acid)
Since base = acid then base/acid = 1; log 1 is zero and pH = pKa.

b. Use the Henderson-Hasselbalch equation above and substitute for base and acid then calculate pH.

Can you explain it better we JUST started this section and im having a hard time with knowing what to put where when it comes to the solutions. what is the acid and what is the base

Thank you :) this helped a lot!

A) Well, if formic acid and sodium formate have equimolar concentration, it means they're in a 1:1 ratio. In this case, we can use the Henderson-Hasselbalch equation, which states that pH = pKa + log([A-]/[HA]). Since we have equal amounts of formic acid and sodium formate, [A-]/[HA] would be 1. So, the pH would be pK3, which is 3.75. Easy peasy, lemon squeezy!

B) Ah, looks like the sodium formate is feeling a bit richer with 10M, while formic acid is feeling a bit poor with only 1M. No worries, my friend! We can still use the Henderson-Hasselbalch equation. So, pH = pKa + log([A-]/[HA]). In this case, [A-]/[HA] would be 10/1, which is equal to 10. So, our equation becomes pH = pK3 + log(10). Now, I don't have a calculator handy, but I'm pretty sure if you do the math, you'll get your pH. Good luck!

To answer both questions, we need to understand the concept of a buffer and how it works.

A buffer is a solution that resists changes in pH when small amounts of acid or base are added to it. Buffers are composed of a weak acid and its conjugate base (or a weak base and its conjugate acid). In this case, the weak acid is formic acid (HCOOH) and its conjugate base is sodium formate (HCOONa).

The pH of a buffer can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

Where:
- pKa is the logarithmic value of the acid dissociation constant (Ka)
- [A-] is the concentration of the conjugate base (sodium formate)
- [HA] is the concentration of the weak acid (formic acid)

Now let's calculate the pH for each scenario:

A) In a buffer with equimolar concentrations of formic acid and sodium formate:
Since the concentrations are equimolar, we can assume that [A-] = [HA].
Therefore, we substitute [A-] = [HA] in the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
= pKa + log(1)
= pKa

Given that pKa = 3.75, the pH of the buffer will be 3.75.

B) In a solution with 10M sodium formate and 1M formic acid:
Substitute the respective concentrations in the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
= pKa + log(10/1)
= pKa + log(10)
= 3.75 + log(10)

Using a calculator, we can find log(10) ≈ 1:
pH = 3.75 + 1
= 4.75

Therefore, the pH of the solution with 10M sodium formate and 1M formic acid will be 4.75.