Chloropropionic acid, CICH2CH2COOH is a weak monoprotic acid with Ka = 7.94 x 10-5 M. Calculate...

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Question

Answer 1

Answer #1

Q1)pKa of acid = -log Ka = -log 7.94x10-5 = 4.1

At equivalence

mmoles of acid = mmoles of base

48.7mL x 0.39M = V x 0.70M

Thus volume of base at equivlaence = 27.13 mL

At equivalence

HA + KOH ----------------------------> KA + H2O

48.7x0.39 27.13x0.70 0 0 initial mmoles

0 0 18.993 --- equilibrium

The concentration of KA = [KA] = mmoles/total volume = 18.993/(48.7+27.13)=0.25 0M

The solution now has only the salt of weak acid with strong base and hence is basic in aquous solution due to anionic hydrolysis.

The pH of such salt is given by

pH = 1/2[pKw +pKa + log C]

= 1/2[14+4.1 + log 0.250]

= 8.749

Answer 2

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