Chapter+12,+Section+3

=Equilibrium Constants=

Key Concepts
Equilibrium Constants (K): > > > pP + qQ where A, B, P and Q are the chemical species and > a, b, p and q are the stoichiometric coefficients,
 * are derived from experimental data
 * are temperature dependent
 * provide a measure of the equilibrium position
 * # if K is large (> 102) products of the reaction are favoured
 * 1) 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 [[image:http://www.ausetute.com.au/images/eqlarrow.gif width="15" height="12" align="center"]]



Solubility Product (Ksp)
> > bB(aq) + cC(aq) > ||^ || > ||^ || [A]a || > 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)
 * refers to an ionic compound dissolving to form ions
 * for the reaction: aA(s) [[image:http://www.ausetute.com.au/images/eqlarrow.gif width="15" height="12" align="center"]]
 * K = || [B]b[C]c ||
 * 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

> 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 > 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
 * Deciding whether a precipitate will form

[|Chapter 12 Worksheet]