the zwitterion form of the amino acid shown below can react with an acid or base

Question

the zwitterion form of the amino acid shown below can react with an acid or base.

one 10.00-ml aliquot of 0.5000m of the zwitterion form shown above was titrated with 0.2500m NaOH and a second 10 ml with 0.1000m HCl.

Ka1 for the amino acid = 5.00 x 10^-3 ; Ka2 = 1.79 x 10^-10

a) what is the pH of the solution after the addition of 20.00 mL of NaOH to the zwitterion solution?

b) what is the pH after the addition of 50.00 mL of HCl to the zwitterion solution?

I need help answering parts a and b. Can you solve this using an ICE box (the method I was taught and am familiar with) and explain (in words) the concept and the steps behind each step with more of the explanation centered around part b? Thank you.

Explanation / Answer

Titration of amino acid

a) with 20 ml of 0.25 M NaOH

moles of amino acid = 0.5 m x 10 ml = 5 mmol

moles of NaOH added = 0.25 M x 20 ml = 5 mmol

moles of neutralized amino acid = 5 mmol

molarity of neutalized amino acid = 5/30 ml = 0.167 M

hydrolysis in water produces OH- ions

let x amount has hydrolyzed

Kb2 = 1 x 10^-14/1.79 x 10^-10 = x^2/0.167

x = [OH-] = 3.05 x 10^-3 M

pOH = -log[OH-] = 2.515

pH = 14 - pOH = 11.485

b) with 50 ml of 0.10 M HCl

moles of amino acid = 0.5 m x 10 ml = 5 mmol

moles of HCl added = 0.1 M x 50 ml = 5 mmol

moles of neutralized amino acid = 5 mmol

molarity of neutalized amino acid = 5/30 ml = 0.167 M

hydrolysis in water produces H3O+ ions

let x amount has hydrolyzed

Kb2 = 5 x 10^-3 = x^2/0.167

x = [H3O+] = 0.029 M

pH = -log[H3O+] = 1.54

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