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Alja [10]
3 years ago
14

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

Chemistry
2 answers:
nevsk [136]3 years ago
7 0
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? 
<span>pH = pKa + log(base)/(acid) </span>
<span>4 = 4 + log (base)/(acid) </span>
<span>(base)/(acid) = 1 </span>
<span>So 50% has been converted.</span>
Bezzdna [24]3 years ago
7 0

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%

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