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Equilibium Constants

Key Concepts

Equilibrium Constants (K):

  • are derived from experimental data

  • are temperature dependent

  • provide a measure of the equilibrium position
  1. if K is large (> 102) products of the reaction are favoured

  2. if K is small, reactants are favoured.

  • remain the same value for a constant temperature even if the equilibrium concentrations of the reactants or products are altered

  • do NOT change in the presence of a catalyst (a catalyst changes the time taken to reach equilibrium but does not alter the equilibrium position or equilibrium constant)

  • For the reaction:
    aA + bB cC + dD

    where A, B, C and D are the chemical species and
    a, b, c and d are the stoichiometric coefficients,

    K = [C]c[D]d

    [A]a[B]b

Some Special Special Cases

Ion-product of Water (Kw)

  • Ion-product of water refers to self-ionization of water:
    H2O H+ + OH-
    K = [H+][OH-]

    [H2O]

  • Since H2O is a pure substance its concentration is a constant at a given temperature and pressure so its concentration is incorporated into the overall equilibrium constant:
    Kw = [H+][OH-]

  • At 25o (298K), Kw = 10-14 M2

  • Since Kw is small, the reactant, H2O, is favoured,
        so very few H+ and OH- ions are present at equilibrium.

  • Calculating [H+] and [OH-] at equilibrium:
        at equilibrium [H+] = [OH-] (from the balanced chemical equation)
        Kw = [H+]2 = 10-14
        [H+] = [OH-] = 10-7 mol L-1

Solubility Product (Ksp)

  • refers to an ionic compound dissolving to form ions

  • for the reaction:
    aA(s) bB(aq) + cC(aq)

    K = [B]b[C]c

    [A]a

  • Since the concentration of the solid is a constant, it is effectively incorporated into the equilibrium constant:
    Ksp = [B]b[C]c

  • Calculating how much solid will dissolve in water
        For example, calculate how much silver bromide will dissolve in 1 L of water
        (Ksp at 25oC is 5.0 x 10-13)

        AgBr(s) Ag+(aq) + Br-(aq)
        Ksp = [Ag+][Br-] = 5.0 x 10-13
        At equilibrium [Ag+] = [Br-] (from the balanced chemical equation)
        5.0 x 10-5 = [Ag+]2
        [Ag+] = [Br-] = 7 x 10-7 mol L-1
        The solubility of AgBr at 25oC is 7 x 10-7 M

  • Deciding whether a precipitate will form
        For example, will a precipitate form if 25.0 mL of 1.4 x 10-9 M NaI and 35.0 mL of 7.9 x 10-7 M AgNO3 are mixed? (Ksp for AgI at 25oC is 8.5 x 10-17)

        Calculate the concentration of I- after mixing:
        M1 = [I-] before mixing = 1.4 x 10-9 M (assuming full dissociaiton of NaI)
        V1 = initial volume = 25.0 mL = 25.0 x 10-3 L
        V2 = final volume after mixing = 25.0 + 35.0 = 60.0 mL = 60.0 x 10-3 L (assuming volumes are additive)
        M2 = [I-] after mixing
        M2 = [I-] after mixing = M1 x V1 ÷ V2 = (1.4 x 10-9 x 25.0 x 10-3) ÷ (60.0 x 10-3) = 5.8 x 10-10 M

        Calculate the concentration of Ag+ after mixing:
        M1 = [Ag+] before mixing = 7.9 x 10-7 M (assuming full dissociaiton of AgNO3)
        V1 = initial volume = 35.0 mL = 35.0 x 10-3 L
        V2 = final volume after mixing = 25.0 + 35.0 = 60.0 mL = 60.0 x 10-3 L (assuming volumes are additive)
        M2 = [Ag+] after mixing
        M2 = [Ag+] after mixing = M1 x V1 ÷ V2 = (7.9 x 10-7 x 35.0 x 10-3) ÷ (60.0 x 10-3) = 4.6 x 10-7

        Calculate the ion product:
        K = [Ag+][I-] = 5.8 x 10-10 x 4.6 x 10-7 = 2.7 x 10-16

        Decide whether a precipitate forms:
        If ion product > Ksp a precipitate will form
        If ion product < Ksp a precipitate will not form
        In this case, 2.7 x 10-16 > Ksp (8.6 x 10-17) so a precipitate will form

Acid Dissociation Constants (Ka)

For the reaction:

HB H+ + B-

Ka = [H+][B-]

[HB]

  • The stronger an acid is, the more it dissociates producing more H+ and B-,
        so a stronger acid has a larger Ka

  • Strong acids completely dissociate in solution so their Ka values approach infinity,
        eg, HCl(aq)

Calculating [H+]
For example: calculate the [H+] in 0.10 mol L-1 HNO2(aq). (Ka = 5.0 x 10-4 at 25oC)

HNO2 H+ + NO2-

let x = moles of HNO2 that dissociate to form H+ and NO2-
  HNO2 H+ + NO2-
initial concentrations (M) 0.10   0   0
Equilibrium concentrations 0.10 - x   x   x

Ka = [H+][NO2-]

[HNO2]
5.0 x 10-4 = [x][x]

[0.10 - x]

Since the acid dissociates only slightly, x will be very small compared to 0.10

so 0.10 - x ~ 0.10
5.0 x 10-4 = x2 ÷ 0.10

x = [H+] = 7.1 x 10-3 mol L-1

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