Question

Determine the pH of each of the following solutions

A)8.1×10⁻³ M phenol

B)9.5×10⁻² M hydroxylamine.

Calculate the percent ionization of propionic acid (C2H5COOH) in solutions of each of the following concentrations (Ka is given in Appendix D in the textbook).

A)0.260 M .

B) 8.05×10⁻² M .

c)1.99×10⁻² M .

Answer 1

a)

phenol pKa = 9.95

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 = 8.1*10^-3 M; then

x^2 + (10^-9.95)x - (8.1*10^-3)*(10^-9.95) = 0

solve for x

x =9.53*10^-7

substitute

[H+] = 0 + 9.53*10^-7= 9.53*10^-7M

pH = -log(H+) = -log(9.53*10^-7) = 6

b)

This is a base in water so, let the base be CH3NH2 = "B" and CH3NH3+ = HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibrium Kb:

Kb = [HB+][OH-]/[B]

initially:

[HB+] = 0

[OH-] = 0

[B] = M

the change

[HB+] = x

[OH-] = x

[B] = - x

in equilibrium

[HB+] = 0 + x

[OH-] = 0 + x

[B] = M - x

Now substitute in Kb

Kb = [HB+][OH-]/[B]

Kb = x*x/(M-x)

x^2 + Kbx - M*Kb = 0

x^2 + (10^-7.97)x - (9.5*10^-2)(10^-7.97) = 0

solve for x

x = 3..189*10^-5

substitute:

[HB+] = 0 + x =  3.189*10^-5M

[OH-] = 0 + x = 3.189*10^-5M

pH = 14 + pOH = 14 + log(3.189*10^-5) = 9.5