Question

Find the identity of a solid, weak acid, HY. The following information is obtained a) Write the dissociation equation for HY in water and its equilibrium expression (Ka). b) A sample of 0.2702 g of HY is titrated and found to require 30.00 mL of 0.10 M KOH solution to reach the equivalence point. Calculate the molar mass of HY c) After 6.00 mL of the base is added, the pH of the solution is 3.25. What is the value for Ka of HY? d) At the equivalence point of the reaction, will the solution be acidic, basic, or neutral? Justify your answer. e) What is the identity of the acid (HY)

Answer 1

a)

the dissociation equation is given by

HY ----> H+ + Y-

and

the equilibrium expression is given by

Ka = [H+] [Y-] / [HY]

b)

we know that

at equivalence point

moles of acid = moles of base

also

moles = molarity x volume (L)

so

Moles of KOH = 0.1 x 30 x 10-3

moles of KOH = 3 x 10-3

so

moles of HY = 3 x 10-3

now

moles = mass / molar mass

so

3 x 10-3 = 0.2702 / molar mass

molar mass = 90 g /mol

so

molar mass of HY is 90 g/mol

c)

now

moles of base added = 0.1 x 6 x 10-3 = 6 x 10-4

moles of HY present = 3 x 10-3

now

the reaction is

HY + KOH ----> KY + H20

from the above reaction

moles of HY reacted = moles of KOH added = 6 x 10-4

moles of HY remaining = 3 x 10-3 - 6 x 10-4 = 2.4 x 10-3

also

moles of KY formed = moles of KOH added = 6 x 10-4

we know that

according to hasselbach henderson equation

pH = pKa + log [ salt / acid ]

pH = pKa + log [ KY / HY]

3.25 = pKa + log [ 6 x 10-4 / 2.4 x 10-3 ]

pKa = 3.852

Ka = 1.4 x 10-4

so

the value of Ka is 1.4 x 10-4

d)

at equivalence point , the solution will be basic

e) molar mass is 90

and

Ka = 1.4 x 10-4

so

the acid (HY ) is lactic acid