Question

M formic acid solution is titrated against 0.1 M NaOH solution. What would be the difference in pH between $\dfrac{1}{5}\& \dfrac{4}{5}$ stages of neutralization of acid?
Option:
(A) 2 log $\dfrac{3}{4}$
(B) 2 log $\dfrac{1}{3}$
(C) 2 log 4
(D) log $\dfrac{1}{3}$

Answer 1

The simplest carboxylic acid is formic acid, also known as methanoic acid. Its chemical formula is HCOOH. It's a crucial step in chemical synthesis that can be seen in nature, most commonly in ants. The term "formic" comes from the Latin word for ant, formica, and refers to its early isolation through ant bodies distillation. Formates are esters, salts, and the anion extracted from formic acid. Methanol is used to make formic acid in industry.

Complete answer:
The Henderson-Hasselbalch equation describes the relationship between acid pH and pKa in aqueous solutions (acid dissociation constant). When the concentrations of the acid and its conjugate base, or the base and the related conjugate acid, are known, this equation can be used to approximate the pH of a buffer solution. The Henderson-Hasselbalch formula is as follows:
${\text{pH}} = {\text{p}}{K_{\text{a}}} + {\log _{10}}\left( {\dfrac{{[{\text{ Base }}]}}{{[{\text{Acid}}]}}} \right)$
Lawrence Joseph Henderson, an American chemist, was the first to develop an equation that could determine the pH value of a given buffer solution. Karl Albert Hasselbalch, a Danish chemist, re-expressed this equation in logarithmic terms. The Henderson-Hasselbalch Equation was created as a result.
Direct methods can be used to quantify the ionisation constants of strong acids and strong bases. However, since the degree of ionisation of weak acids and bases is very limited, the same techniques cannot be used for them (weak acids and bases hardly ionize).
As we know
${\text{pH}} = {\text{p}}{K_{\text{a}}} + {\log _{10}}\left( {\dfrac{{[{\text{ Base }}]}}{{[{\text{Acid}}]}}} \right)$
Then at first stage
$\dfrac{1}{5}$= This suggests that out of a total of five, one is the base and the other four are the acids. So the HCOOH concentration is 80% and the $HCO{O^ - }$concentration is 20%.
${\text{pH}} = {\text{p}}{K_{\text{a}}} + {\log _{10}}\left( {\dfrac{1}{4}} \right)$-- (1)
$\dfrac{4}{5}$= This suggests that out of a total of five, one is the base and the other four are the acids. So the HCOOH concentration is 20% and the $HCO{O^ - }$concentration is 80%.
${\text{pH}} = {\text{p}}{K_{\text{a}}} + {\log _{10}}\left( {\dfrac{4}{1}} \right)$--(2)
When we subtract (2) and (1)
${\text{p}}{{\text{H}}_2} - {\text{p}}{{\text{H}}_1} = \log 4 - ( - \log 4) = 2\log 4.$
Hence 2 log 4 is correct. Option (C) is correct.

Note:
Since it predicts that the concentration of the acid and its conjugate base at chemical equilibrium will remain the same as the formal concentration, the Henderson-Hasselbalch equation fails to determine precise values for strong acids and strong bases. The Henderson-Hasselbalch equation struggles to have correct pH values for highly dilute buffer solutions because it ignores water's self-dissociation.