Question

Sodium acetate (NaOAc) is the sodium salt of the conjugate base of acetic acid (HOAc). Although NaOAc is a strong electrolyte, the acetate ion itself (OAc−) is a weak base. Assume pKa = 4.76 for HOAc.a. (1 pt) Calculate the pH of a 50.0 mM HOAc solution. Using a 5% criterion, exp licitly check any approximations you make to ensure they are valid.b.(1 pt) Calculate the pH of a 50.0 mM NaOAc solution. Again, check your assumptions.

Answer 1

The buffer equation for the acetic acid:

pH = pKa + log(OAc-/HOAc)

a)

pH of a 50mM HOAc

First, assume the acid:

HF

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = 50*10^-3 M; then

x^2 + (1.8*10^-5)x - (50*10^-3)*(1.8*10^-5) = 0

solve for x

x =9.39*10^-4

substitute

[H+] = 0 + 9.39*10^-4= 9.39*10^-4M

[A-] = 0 + 9.39*10^-4= 9.39*10^-4 M

[HA] = M - x = 50*10^-3-9.39*10^-4= 0.049061M

pH = -log(H+) = -log(9.39*10^-4) = 3.027

b)

find pH of a 50 mM

expect hydrolysis

CH3COO-(aq) + H2O(l) <->CH3COOH+ OH-(aq)

Let HA --> CH2OH and A- = CH2O- 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^-10 = x*x/(M-x)

solve for x

x^2 + (5.55*10^-10)*x - (50*10^-3)(5.55*10^-10) = 0

solve for x with quadratic equation

x = OH- =5.26*10^-6

[OH-]  =5.26*10^-6

pOH = -log(OH-) = -log(5.26*10^-6 = 5.27901

pH = 14-5.27901= 8.72

pH = 8.72