Question

what is the fraction of association for the following potassium propionate solutions? Ignore activities. The ka of propanoic acid. 1.34*10^-5.

a. 4.00*10^-1 m k(c2h5co2)

b. 4.00*10^-2 m k(c2h5co2)

c. 4.00*10^-10 m k(c2h5co2)

Answer 1

C2H5CO2-(aq) + H2O(l) <--------> C2H5CO2H(aq) + OH-(aq)
Kb = [OH-]*[C2H5CO2H]/[C2H5CO2-]
If the portion of reactant i.e. x moles of C2H5CO2-undergo reaction with H2O, that will produce x moles of C2H5CO2H and x moles of OH-,
Kb = x²/(A - x)

Kb = 10^-14/Ka = 7.46*10^-19 = x²/(A - x)

Solving for x in each case
(a) A = 4.00*10^-1, then 7.46*10^-19 = x2/ (4.00*10^-1 -x)

(4.00*10^-1 -x) x 7.46*10^-19 =x²

x² + 7.46*10^-19x- 4.00*10^-1 * 7.46*10^-19 =o

whichc is quardatic equation that can be solved using the equation