Equilibrium ❮ 229

(^) The strong base will control the pH.
[OH-] = 0.0015 mol/0.155 L = 9.7 × 10 -^3 M
The stoichiometry part of the problem is finished.
Since this is now a solution of a strong base, it is now a simple pOH/pH problem.
pOH = - log 9.7 × 10 -^3 = 2.01
pH = 14.00 - pOH = 14.00 - 2.01 = 11.99
f. 75.00 mL. Since both an acid and a base are present (and they are not conjugates), this
must be a stoichiometry problem. Stoichiometry requires a balanced chemical equation
and moles.
+→++

HNO NaOH Na+ NO HO
Base:
0.300mol
1,000 mL
75.00ML 0.0225mol
22

• 2

Based on the stoichiometry of the problem, and on the moles of acid and base, the

acid is now the limiting reagent.

+→++

−− ++

HNONaOHNaN+−OHO

init. 0.0150 0.0225mol0 0

react. 0.0150 0.0150 0.0150 0.0150

final 0.0000 0.0075 — 0.0150

222

The strong base will control the pH.

[OH-] = 0.0075 mol/0.175 L = 4.3 × 10 -^2 M

The stoichiometry part of the problem is finished.

Since this is now a solution of a strong base, it is now a simple pOH/pH problem.

pOH = - log 4.3 × 10 -^2 = 1.37

pH = 14.00 - pOH = 14.00 - 1.37 = 12.63

Solubility Equilibria

Many salts are soluble in water, but some are only slightly soluble. These salts, when placed

in water, quickly reach their solubility limit and the ions establish an equilibrium system

with the undissolved solid. For example, PbSO 4 , when dissolved in water, establishes the

following equilibrium:

PbSO( 4 s)Pb2+(aq)+SO 42 −(aq)

The equilibrium constant expression for systems of slightly soluble salts is called the

solubility product constant, Ksp. It is the product of the ionic concentrations, each one

raised to the power of the coefficient in the balanced chemical equation. It contains no

denominator since the concentration of a solid is, by convention, 1 and does not appear in

the equilibrium constant expressions. (Some textbooks will say that the concentrations of