Question

100 mL of 1.0 M formic acid (HCOOH) is titrated with 1.0 M sodium hydroxide (NaOh).

The Approximate pKa is 4 At the pKa, what fraction of the carboxyl group will have been converted to COO-?

Answer 1

At the pKa, 100% of the carboxyl group will have been converted to COO-.

Step-by-step explanation:

The fraction of the carboxyl group that will have been converted to COO- at the pKa can be calculated using the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is given by:

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

In this case, formic acid (HCOOH) acts as the weak acid and its conjugate base, formate (HCOO-), acts as the weak base. At the pKa, the concentration of the weak acid and weak base will be equal. Therefore, the ratio [A-]/[HA] is 1. By substituting this value into the Henderson-Hasselbalch equation, we get:

pKa = pH

So, at the pKa, 100% of the carboxyl group will have been converted to COO-.

Answer 2

Fraction of carboxyl group that will have been converted to COO- = 50%

Step-by-step explanation:

pKa of formic acid = 4

At pKa, titration is at equivalence point.

At equivalence point, number of moles of HCOOH undissociated is equal to number of moles of dissociated
`(HCOO^-)`

`(HCOOH)_(eq)= (HCOO^-)_(eq)`

Thus, fraction of carboxyl group that will have been converted to COO-:

`([HCOO^-]_(eq))/([HCOO^-]_(eq)+[HCOOH]_(eq)) * 100 = ([HCOO^-]_(eq))/([HCOO^-]_(eq)+[HCOO^-]_(eq))`

`([HCOO^-]_(eq))/(2[HCOO^-]_(eq)) * 100 = 50 \%`

fraction of carboxyl group that will have been converted to COO- = 50%