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Wize High School Grade 12 Chemistry Textbook > Acids and Bases

Percent Dissociation

Applications of Acid and Base EquationsPrevious Section
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Percent Ionization or Dissociation


p=(concentration of acid ionized)(concentration of acid solute)x100%p=\frac{\left(concentration\ of\ acid\ ionized\right)}{\left(concentration\ of\ acid\ solute\right)}x100\%
p=percent ionization

For the general weak acid equation:

HA(aq) ⇌ H+(aq) + A-(aq)

p=[H+][HA]x100\boxed{p=\frac{\left[H+\right]}{\left[HA\right]}x100}

  • Do you think a weak acid would have a high or low percent ionization?
    Low
  • Do you think a strong acid would have a high or low percent ionization?
    High
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Example: Percent Dissociation

A 0.150 M solution of acetic acid (shorthand formula = HAc) is found to be 1.086% dissociated. What is the Ka?


The equation for the dissociation of HAc is:
HAc ⇌ H+ + Ac¯

The Ka expression is:

Ka=[H+][Ac−][HAc]Ka=\frac{\left[H^+\right]\left[Ac^-\right]}{\left[HAc\right]}

The concentration we are given for HAc is the initial concentration but we need to plug in the equilibrium concentration into our Ka expression.

1) Draw out an ICE Table so we can determine equilibrium concentrations


i 0.15M 0 0
C -x +x +x E 0.15-x x x

2) We find [H+] using the concentration and the percent dissociation:


Use this equation and solve for [H+]:
p=1.086

p=[H+][HA]x100\boxed{p=\frac{\left[H+\right]}{\left[HA\right]}x100}

p[HA]100=[H+]\frac{p[HA]}{100}=\left[H+\right]

1.068[0.15M]100=[H+]\frac{1.068\left[0.15M\right]}{100}=\left[H+\right]

H+=1.629 x 10¯3 M


3) Solve for [Ac¯]

x= [Ac¯] = [H+] = 1.629 x 10¯3 M

4) Solve for [HAc]

[HAc]=0.15-x
=0.15-1.629x10-3
=0.148371

4) Now, we insert into the Ka expression and solve:


Ka=[H+][Ac−][HAc]Ka=\frac{\left[H^+\right]\left[Ac^-\right]}{\left[HAc\right]}

Ka =[1.629x10−3]20.148371Ka\ =\frac{\left[1.629x10-3\right]2}{0.148371}

Ka= 1.8x10-5