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Wize University Chemistry Textbook > Acids and Bases

Percent Dissociation

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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
Extra Practice
In a 0.25 M solution, a weak acid is 3.0% dissociated. Calculate [H3O+], pH and pOH of the solution
In a 0.25 M solution, a weak acid is 3.0% dissociated. Calculate [H3O+], pH and pOH of the solution
Weak Acids and Bases
Calculate the percent dissociation for the following solutions.
a) A 0.223 M aqueous solution of HF.
b) A 1.70 x 10-2 M aqueous solution of NaF.
Weak Acids and Bases
Which equation best describes the dissociation of hydrofluoric acid in water?
Weak Acids and Bases
Calculate the percent dissociation for the following solutions.
a) A 0.444 M aqueous solution of HCN (Ka = 6.2 x 10-10).
b) A 3.50 x 10-2 M aqueous solution of NaCN.
Acids and Bases: pH based on % dissociation
In a 0.25 M solution, a weak acid is 3.0% dissociated. Calculate [H3O+], pH and pOH of the solution
If an answer is an exponent such as 4.3x10-7, input the answer as: 4.3e-7
Le Chatelier's Principle: Percent dissociation
A solution made by dissolving 0.1 mol of acetic acid in 1.0 L of water is 4.1% dissociated. Which of the modifications below would cause the % dissociation of the acetic acid in the container to increase? (Ka for acetic acid is 1.8×10-5.)