Question

2a. Write all three hydrolysis reactions and acid ionization expressions
(KA1, KA2, and KA3) for weak acid L-cysteine (H3Cys).
Include all ionic charges. Use these expressions in parts b, c, and d. (9 pts)

b. Use an eqm table to determine [H3O+1] of 2.85 M H3Cys if KA1 = 1.20 × 10−2. Ignore effects of KA2 and KA3. Note that [H3Cys]/KA1 > 100. (4 pts)

c. Determine [H2Cys−1] from your equilibrium table above. (4 pts)

d. Use an equilibrium table to determine [HCys −2] for the same solution where KA2 = 4.27 × 10−9. Use concentration values from parts b and c. (4 pts)

e. Use an equilibrium table to determine [Cys −3] for the same solution where KA3 = 2.00 × 10−11. Use concentration values from the parts b, c, and d.
(4 pts)

Answer 1

2.(a).

First ionization expression is :

H₃Cys ...............+................H₂O <---------------> H₃O⁺ ........................+..................H₂Cys^-1

Second ionization expression is :

H₂Cys^-1 ...............+................H₂O <---------------> H₃O⁺ ........................+..................HCys^-2

and

Third ionization expression is :

HCys^-2 ...............+................H₂O <---------------> H₃O⁺ ........................+..................Cys^-3

(b). ICE table of H₃Cys is :

.............................H₃Cys ...............+................H₂O <---------------> H₃O⁺ ........................+..................H₂Cys^-1

Initial....................2.85 M...................................................................0.0 M..............................................0.0 M

Change.................-y..........................................................................+y.........................................................+y

Equilibrium..........(2.85-y) M.............................................................y M.......................................................y M

Where,

y = Amount dissociated per mole

Now,

Expression of Equilibrium constant i.e. K_a1(which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

K_a1 = [H₃O⁺].[H₂Cys^-1] / [H₃Cys]

1.20 x 10^-2 = y² / (2.85-y)

y² + 1.20 x 10^-2y - 0.0342 = 0

On solving this equation

y = 0.179

Therefore,

Concentration of H3O+ = [H3O+] = 0.179 M

(c).

Also,

Equilibrium concentration of H2Cys-1 = [H2Cys-1] = 0.179 M

(d).

ICE table of H₂Cys^-1 is :

.............................H₂Cys^-1 ...............+................H₂O <---------------> H₃O⁺ ........................+..................HCys^-2

Initial....................0.179 M...................................................................0.179 M..............................................0.0 M

Change.................-y..........................................................................+y.........................................................+y

Equilibrium..........(0.179 -y) M.......................................................(0.179 +y) M..........................................y M

Where,

y = Amount dissociated per mole

Now,

Expression of Equilibrium constant i.e. K_a2 (which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

K_a2 = [H₃O⁺].[HCys^-2] / [H₂Cys^-1]

4.27 x 10^-9 = y(0.179 +y) / (0.179-y)

As, y <<<0.179, gthen neglect y as compare to 0.179

So,

y = 4.27 x 10^-9  

Therefore,

Equilibrium concentration of HCys-2 = [HCys-2] = 4.27 x 10-9 M

(e).

ICE table of HCys^-2 is :

.............................HCys^-2 ...............+................H₂O <---------------> H₃O⁺ ........................+..................Cys^-3

Initial...................4.27 x 10^-9 M......................................................0.179 M..............................................0.0 M

Change.................-y..........................................................................+y.........................................................+y

Equilibrium..........(4.27 x 10^-9-y) M....................................(0.179 +y) M..........................................y M

Where,

y = Amount dissociated per mole

Now,

Expression of Equilibrium constant i.e. K_a3 (which is equal to the product of the molar concentration of products divided by product of the molar concentration of reactants raise to power their stoichiometric coefficient when reaction is at equilibrium stage).

K_a3 = [H₃O⁺].[Cys^-3] / [HCys^-2]

2.00 x 10^-11 = y(0.179 +y) / (4.27 x 10^-9-y)

As, y <<<0.179, gthen neglect y as compare to 0.179

So,

2.00 x 10^-11 = y(0.179) / (4.27 x 10^-9-y)

2.00 x 10^-11(4.27 x 10^-9-y) = 0.179y

8.54 x 10^-20 - 2.00 x 10^-11 y = 0.179y

0.179y + 2.00 x 10^-11 y = 8.54 x 10^-20

0.179y = 8.54 x 10^-20

y = 8.54 x 10^-20 / 0.179

y = 4.77 x 10^-19  

Therefore,

Equilibrium concentration of Cys-3 = [Cys-3] = 4.77 x 10-19 M