Step 2: Write an equilibrium expression.
K.
NHNH OH
b 18 10(^45)
3
==#
+67 7
@A AStep 3: Set up a chart showing initial concentrations of the species and final concentrations at
equilibrium (the Cl–does not contribute to the pH). Let xrepresent the portion of NH 3 that
eventually converts to NH 4 +. xwill also represent the amount by which the concentration of
NH 4 +increases.
Species Initial Concentration Final Concentration
NH 3 0.35 M 0.35 −x
NH 4 + 0.20 M 0.20 + x
OH– ~0 M* ~x
*Because NH 3 is a weak base and you are using a relatively weak solution (0.35 M), for calculating purposes you can essentially claim that
OHinit–is 0. Further, because [OH–] and NH 4 +at equilibrium are in a 1: 1 molar ratio, the concentration of OH–will increase by the amount x.Step 4: Substitute these chart values into the equilibrium expression.
.
..()
K
xxx
NHNH OH
18 10
035020
b(^45)
3
===#
++^^
hh
67 7
@A ASince, <<<. ,
.
.
KK.x
020
035020
bb==18 10# -^5
^^^
hhh
could be used to solve for x. Therefore C is thecorrect expression, but D could be used to solve for the answer.
Note: If this question were in the free-response section (calculators allowed), and if the
question asked you to solve for the pH of the buffer, you would need to solve for x using
the quadratic equation. x would be 3.2 × 10 –5. Because [OH–] ≈3.2 × 10 –5and pOH = -log
[OH–], pOH = 4.5. pH = 14.0 −pOH = 14.0 −4.5 = 9.5.- Copper(II) iodate has a solubility of 3.3 × 10 –3M at 25°C. Calculate its Kspvalue.
A. 1.4 × 10 –7
B. 1.1 × 10 –5
C. 3.3 × 10 –3
D. 5.1 × 10 –1
E. 3.3 × 103Answer: A
Equilibrium