What is the PH of an aqueous solution consisting of 0.070 M formic acid (HCOOH) and 0.070 M sodium formate (HCOONa)?

Use the Henderson-Hasselbalch equation.

pH = pKa + log (base)/(acid)
pH = ?
pKa = look up
(base) = 0.07
(acid) = 0.07

pKa = 7.6 for HCOOH

so pH= 7.6 + log [.07]/[.07}
pH = 7.6

is this correct or at least moving in the right direction?

Some is correct. I don't know how your calculator works (calculators aren't standard) but I can tell you how to proceed.

First, where did you get the pKa for formic acid? I looked it up on the web and found 3.77. You can get slightly different numbers depending upon where you look but that's fiarly close. The Ka is about 1.7E-4. So
pH = pKa + log (base)/(acid_
pH = 3.77 + log (0.07/0.07)
pH = 3.77 + log 1
and log 1 I thought everyone knew is zero and you really don't need a calculator for that.
Then pH = 3.77 + 0 = 3.77. Right?

To find the pH of the aqueous solution consisting of 0.070 M formic acid (HCOOH) and 0.070 M sodium formate (HCOONa), we need to consider the acid-base equilibrium between formic acid and its conjugate base, formate ion.

Formic acid (HCOOH) can donate a proton (H+) and act as an acid, whereas formate ion (HCOO-) can accept a proton (H+) and behave as a base. The equilibrium equation for this acid-base reaction is as follows:

HCOOH ⇌ H+ + HCOO-

Since both formic acid and sodium formate are present in the solution, we have a buffer system. A buffer resists changes in pH when small amounts of acid or base are added to it.

To calculate the pH of the buffer solution, we use the Henderson-Hasselbalch equation:

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

Where:
pH = the desired pH
pKa = the negative logarithm of the acid dissociation constant of the acid (HCOOH)
[A-] = concentration of the conjugate base (HCOO-)
[HA] = concentration of the acid (HCOOH)

The pKa value for formic acid (HCOOH) is approximately 3.75.

First, we need to calculate the ratio of [A-]/[HA]:

[A-]/[HA] = (0.070 M)/(0.070 M) = 1

Substituting the values into the Henderson-Hasselbalch equation:

pH = 3.75 + log (1) = 3.75

Therefore, the pH of the aqueous solution consisting of 0.070 M formic acid (HCOOH) and 0.070 M sodium formate (HCOONa) is approximately 3.75.

I am not really sure how the find the

pKa. Do I need to find the conjugate acid -base pair first?

pKa = -log Ka

I am not understand how to use log on my calculate, so no answer is calculating.

can you explain how to use log?