Skip to main content
Chemistry LibreTexts

Solutions

  • Page ID
    11056
  • 1. Calculate the equilibrium concentrations and pH for a 0.20 M proprionic acid solution.


    Complete the equilibrium and express the equilibrium concentrations in terms of a single unknown 'x'.

    HC3H5O2 (aq)
    H+(aq)
    + C3H5O2- (aq)
    initial concentration
    0.2 M
    0
    0
    change
    - x
    + x
    + x
    equilibrium concentration
    0.2 - x
    x
    x

    Write the equilibrium expression (Ka) and express the Ka in terms of the equilibrium concentrations.

    [H+][C3H5O2-]
    (x)(x)
    Ka =
    _______________
    =
    ___________
    = 1.3 x 10-5
    [HC3H5O2]
    (0.2 - x)
    (from table)

    Solve for x; since the initial concentration is greater than 0.1 and the Ka less than 10-4, we can drop the '- x' term.

    x2
    solving for 'x' gives
    _______
    = 1.3 x 10-5
    x = 1.61 x 10-3
    0.2

    Solve for the equilibrium concentrations by substituting the value of 'x' into the equilibrium concentrations: [H+] = [C3H5O2-] = x, [HC3H5O2] = 0.2 - x

    [H+] = [C3H5O2-] = 1.61 x 10-3 M, [HC3H5O2] = 1.98 x 10-1 M;

    pH = - log [H+] = - log(1.61 x 10-3)

    pH = 2.79

    2. Calculate the equilibrium concentrations and pH for a 0.20 M carbonic acid solution.
    Compare the answer to problem 1. Note which has the higher Ka and lower pH.
    This is done the same way as problem 1, using a Ka value of 4.2 x 10-7

    H2CO3 (aq)
    H+(aq)
    + HCO3- (aq)
    initial concentration
    0.2 M
    0
    0
    change
    - x
    + x
    + x
    equilibrium concentration
    0.2 - x
    x
    x


    [H+][HCO3-]
    (x)(x)
    Ka =
    _______________
    =
    ___________
    = 4.2 x 10-7
    [H2CO3]
    (0.2 - x)
    (from table)


    x2
    solving for 'x' gives
    _______
    = 4.2 x 10-7
    x = 2.90 x 10-4
    0.2

    [H+] = [HCO3-] = 2.90 x 10-4 M, [H2CO3] = 2.0 x 10-1 M; pH = 3.54

    The higher the Ka, the lower the pH for the same concentration of weak acid. The higher Ka value tells us we have more dissociation of the weak acid, giving a greater [H+].

    3. Calculate the Ka for a 0.3 M solution of HA (weak acid) if the pH = 3.65
    First calculate the [H+] concentration from the pH
    [H+] = 10-pH; [H+] = 10-3.65; [H+] = 2.24 x 10-4

    since 1:1 mole ratio

    [H+] = [A-]; [HA] = 0.3 - [H+]

    Substitute into the Ka expression

    [H+][A-]
    (2.24 x 10-4)(2.24 x 10-4)
    Ka =
    ________
    =
    ________________________
    [HA]
    (0.3 - 2.24 x 10-4)

    Ka= 1.67 x 10-7

    4. Calculate the pH for a 0.2 M pyridine solution.

    C5H5N (aq)
    OH-(aq)
    + C5H5NH+ (aq)
    initial concentration
    0.2 M
    0
    0
    change
    - x
    + x
    + x
    equilibrium concentration
    0.2 - x
    x
    x


    [OH-][C5H5NH+]
    (x)(x)
    Kb =
    ________________
    =
    ___________
    = 2.0 x 10-9

    [C5H5N]

    (0.2 - x)
    (from table)

    Solve for 'x'

    x2
    solving for 'x' gives
    _______
    = 2.0 x 10-9
    x = 2.0 x 10-5
    0.2

    [OH-] = [C5H5NH+] = 2.0 x 10-5 M, [C5H5N] = 2.0 x 10-1 M

    Using the following equations: pOH = - log [OH-]; pH + pOH = 14

    pOH = 4.70, pH = 9.30

    5. Calculate the pH of a 0.34 M HCl solution.

    pH = 0.47, strong acid, 100 % ionization; 1:1 mole ratio, [HCl] = [H+]