Question

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.

Determine [H₂Cys^-1] from your equilibrium table above.

Answer 1

 

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