Question

Consider a solution of 0.0500 M isoleucine. If we call the fully protonated form H2A+, find the pH of a solution of: (a) 0.0500 M H2ACl, (b) 0.0500 M HA, and (c) 0.0500 M NaA. For the intermediate form, HA, calculate the pH using any reasonable approximations.

pKa1 = 2.31

pKa2 = 9.60

Answer 1

Solution :-

Lets calculate the ka from the given pka values

Pka= -log ka

Ka = 10^(-pka)

Ka1= 10^(-2.31)

       = 0.0049

Ka2=10^(-9.60)

       = 2.51*10^-10

Part a) Calculating the pH for the 0.0500M H2ACl

      H2A^+   + H2O ----- > H3O^+ + HA

0.0500 M                              0               0

-x                                            +x            +x

0.0500-x                                x               x

Ka1=[H3O^+][HA]/[H2A^+]

0.0049 = [x][x]/[0.0500-x]

0.0049*[0.0500-x] = x^2

0.000245 – 0.0049x =x^2

-x^2 – 0.0049x + 0.000245 =0

solving for the x using the quadratic equation we get

x=0.0134 m= [H3O+]

the ka2 is very small therefore we neglect the second dissociation.

pH= -log [H3O^+]

pH= -log[0.0134]

pH= 1.87

part b) calculating the pH of 0.0500 M HA

     HA + H2O ---- > H3O^+ + A^-

0.0500 M                        0            0

-x                                       +x        +x

0.0500-x                            x           x

Ka2=[H3O^+][A^-]/[HA]

2.51*10^-10 = [x][x]/[0.0500-x]

Since the ka is very small therefore we can neglect the x from denominator

2.51*10^-10 = [x][x]/[0.0500]

2.51*10^-10 *0.0500 = x^2

1.26*10^-11 = x^2

Taking square root on both sides we get

3.55*10^-6 = x = [H3O^+]

pH= -log [H3O^+]

pH= -log [3.55*10^-6]

pH= 5.45

part c) Calculating the pH fo 0.0500 M NaA

A^-   + H2O   ----- > HA + OH^-

0.0500                        0           0

-x                                +x           +x

0.0500-x                     x            x

Kb =[HA][OH^-]/[A^-]

Kb= kw/ka2

Kb= 1*10^-14 / 2.51*10^-10

    = 3.98*10^-5

3.98*10^-5 = [x][x]/[0.0500-x]

Neglect x from denominator

3.98*10^-5 = x^2/ 0.0500

3.98*10^-5 * 0.0500 = x^2

1.99*10^-6 = x^2

Taking square root on both sides

1.41*10^-3 = x =[OH^-]

pOH= -log [OH^-]

pOH= -log[1.41*10^-3]

pOH= 2.85

pH = 14- pOH

pH= 14 – 2.85

pH= 11.15