Question

0.1M Formic Acid solution is titrated against 0.1 M NaOH solution. What would be the difference in pH between 1/5 and 4/5 stages of neutralization of acid?

Answer 1

See below

Explanation:

Formic acid has a pKa of 3.75. Formic acid falls apart into formate ion and `H^+` ion. Formic acid is `HCOOH` (acid), Formate ion is `HCOO^-` (base). You have to use the Henderson-Hasellbalch equation for this. `pH=pka+log("base"/"acid")`

1/5: This means that out of 5, 1 is the base, 4 are the acid. So `HCOOH` concentration is 80%, `HCOO^-` concentration is 20%
`HCOOH` =0.80.1=0.08
`HCOO^-` =0.20.1=0.02
Plug these into the equation, and you get pH= 3.15

4/5: This means that you have 80% `HCOO^-`, 20% `HCOOH`. So you basically plug the same numbers in as above, but the LOG numbers are flipped. pH=4.35

since Log(Concentration/Concentration) has (mol/L)/(mol/L), the liters are the same, so they cancel out. You don't really have to worry about the volume of the combined solution, since volume cancels.