Question

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

Answer 1

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Do you mean at a pH = pKa, what is the fraction? 
pH = pKa + log(base)/(acid) 
4 = 4 + log (base)/(acid) 
(base)/(acid) = 1 
So 50% has been converted.

Answer 2

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%