Question

What is the ph of 0.50 M NaC2H3O2?

Answer 1

NaC2H3O2 = NA+ + C2H3O2-

expect hydrolysis

C2H3O2-(aq) + H2O(l) <->C2H3O2H + OH-(aq)

Let HA --> C2H3O2H and A- = C2H3O2- for simplicity

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

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

Ka can be calculated as follows:

Kb = Kw/Ka = (10^-14)/(1.8*10^-5) = 5.55*10^-10

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

substitute in Kb expression

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

5.55*10^-11 = x*x/(M-x)

solve for x

x^2 + Kb*x - M*Kb = 0

solve for x with quadratic equation

x = OH- =1.665*10^-5

[OH-]  =1.665*10^-5

pOH = -log(OH-) = -log(1.665*10^-5 = 4.7785

pH = 14-4.7785= 9.2215

pH = 9.2215