Answered

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-?

Verified Answer :

Answer:

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

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-:

\(\frac{[HCOO^-]_{eq}}{[HCOO^-]_{eq}+[HCOOH]_{eq}} \times 100 = \frac{[HCOO^-]_{eq}}{[HCOO^-]_{eq}+[HCOO^-]_{eq}}\)

\(\frac{[HCOO^-]_{eq}}{2[HCOO^-]_{eq}} \times 100 = 50 \%\)

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