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 [tex](HCOO^-)[/tex]
[tex](HCOOH)_{eq}= (HCOO^-)_{eq}[/tex]
Thus, fraction of carboxyl group that will have been converted to COO-:
[tex]\frac{[HCOO^-]_{eq}}{[HCOO^-]_{eq}+[HCOOH]_{eq}} \times 100 = \frac{[HCOO^-]_{eq}}{[HCOO^-]_{eq}+[HCOO^-]_{eq}}[/tex]
[tex]\frac{[HCOO^-]_{eq}}{2[HCOO^-]_{eq}} \times 100 = 50 \%[/tex]
fraction of carboxyl group that will have been converted to COO- = 50%
