CHM 2046C Module 12
Name:
____ Sample Answers___
Part N: Predicting Precipitation Reactions 10 points
If the concentration of the
strontium ion, Sr 2+, is 2.5 x 10 -4
M, does precipitation of SrSO4 occur
when enough of the soluble salt Na2SO4
is added to make the solution, 2.5 x 10 -4 M? (Ksp
for SrSO4 is 3.4 x 10 -7).
(b) If no
precipitate forms, what must the concentration of the Sodium Sulfate be in
order for precipitation to begin?
Use the experimental
ion concentrations to calculate the reaction quotient, Q. Then compare the Q
and the Ksp to decide if the system is at
equilibrium (when Q = Ksp).
For Strontium Sulfate:
SrSO4 (s)
↔ Sr2+ (aq)
+ SO42- (aq)
For Sodium Sulfate ( a very
soluble salt-all Sodium salts are soluble):
Na2SO4 (s)
→ 2Na1+ (aq)
+ SO42- (aq)
The Sodium ion is a spectator and is
unreactive, while the concentration of the sulfate
ion is 2.5 x 10 -4 M and provides the Sulfate common Ion, so that
Q = [Sr2+] [SO42-]
= [2.5 x 10-4][ 2.5 x 10-4 ] = 6.25 x 10-8
The Ksp for Strontium Sulfate is 3.4 x 10 -7
Therefore the Ksp > Q
or 3.4 x 10 -7 > 6.25 x
10-8 via Case #2 above the
solution is not saturdated and NO precipitate will form
(b)
If no precipitate forms, what must the concentration of the Sodium Sulfate be
in order for precipitation to begin?
Ksp = [Sr2+] [SO42-]
3.4 x 10 -7 = [2.5 x 10-4][ SO42-]
when [ SO42-] = 1.36 x10-3 M makes Q = Ksp and creates a saturated solution, Any addition amount of Sodium Sulfate added, which cause precipitation of Strontium Sulfate.
Reference McMurry Section 15.13 pages 629-631 which develops the
concepts in Sections 5.14 and 15.15 and Qualitative Analysis. See worked
example 15.14 page 630. Try problems 15.19 and 15.30
on pages 630-631. Also try probelms 15.130-15.134
page 642.
Additional worked example
from another textbook:
