TBL 2 #2

[Lauren Leigh Smith]

2.  A monoamino dicarboxylic acid has the following pKa values:  pKa = 2.1, 3.7, 10.5.  Determine

(a) The isoelectric point, (b) Thenumber of  mmoles of HCl required to titrate 100 ml of the 0.2

M solution of the monosodium salt to the isoelectric point.  (Hint: The monosodium salt in water

has a net charge of -1.0), (c) The pH of a 0.1 M solution of the hydrochloride salt.  (Hint:  Weak

Acid that bears a net charge of +1.0), and (d) The pH where the net charge of the amino acid

molecules is -0.5.  

a.  pI: 2.1 + 3.7/ 2 = 2.9

 

b.  number of mmoles of HCL required to titrate

i.  NaX + HCl  Na + Cl + HX (titrate completely pI)

1.  equivalent amount of HCl to NaX

2.  .2M x .1L = .02mol = 20 mmoles

ii.  Normality: number of H+ for acids and OH- for bases

 

c.  pH of 0.1 M solution of monosodium salt

i.  0.1M, must determine [h+]

ii.  Since we have pka, we know ka

1.  pka = -log (ka)

a.  Ka = 10-2

iii.  Ka = (h+) (a-)/ (ha)

iv.  Ka= x2/ .1

v.  10-2 = x2/ .1M

vi.  x = .0316 = [H+]

vii.  pH = - log [H+]

viii.  pH = 1.5

d.  pH where net charge of AA is -0.5

i.  pKa of 2.1, 50% of unit is dissociated

  net charge +.5

ii.  bw 3.7 and 10.1, net charge is -1

iii.  bw 2.1 and 3.7, it’s heading to ) 

  pI = 2.9 

iv.  at 3.7, lost carboxyl alpha h+ 

  two species, COO-, COO- and NH3+ (charge is -1)

a.  other one, COOH, NH3+, COO- (charge is 0)