TSU Measurement of Ksp values & Determining Concentration of Unknown NaCl Solution Lab

Measurement of Ksp values andDetermining Concentration of an Unknown NaCl Solution
Introduction
Last semester in General Chemistry I, you learned solubility rules to identify combinations of ions that
produce insoluble compounds. In that context, solubility was binary – either a substance was soluble or
not. Those insoluble compounds are more accurately described as poorly soluble, requiring very low
(but measurable) concentrations to precipitate from solution. The solubility of any ionic compound can
be represented as an equilibrium between the pure solid and the dissociated aqueous ions. When the
solid is placed on the reactants side of the equilibrium, the equilibrium constant is known as the
solubility product, Ksp. Some examples are shown below:
⇌
Ba(NO3)2 (s)
BaCO3 (s)
⇌
Ba2+ (aq)
Ba2+ (aq)
+
+
2 NO3– (aq)
Ksp1 = [𝐵𝑎2+ ][𝑁𝑂3 − ]2 = 4.64 × 10−3
CO32– (aq)
Ksp1 = [𝐵𝑎2+ ][𝐶𝑂3 2− ] = 2.58 × 10−9
The magnitude of Ksp gives some idea of how soluble a compound is – poorly soluble compounds have
very small Ksp values. Be careful, though – here be dragons. Because Ba(NO3)2 and BaCO3 produce
different numbers of ions and have different exponents in the equilibrium expression, their Ksp values
are not directly comparable. We can use an ICE table and Ksp for each compound to calculate its molar
solubility, the molarity of a saturated solution.
Initial
Change
Equilibrium
Ba(NO3)2 (s)
–
⇌
Ba2+ (aq)
0
+x
x
+
2 NO3– (aq)
0
+2x
2x
Ksp = [Ba2+][NO3-]2
4.64 × 10−3 = (𝑥)(2𝑥)2
4.64 × 10−3 = 4𝑥 3
x = 0.105 M (molar solubility) → 27.4 g/L
For a generic ionic compound (MaXb) dissolved in water, Ksp = [ax]a[bx]b. For example, Al2O3 would have
the expression Ksp = (2x)2(3x)3. The 1:1 ratio of ions in BaCO3 gives the expression Ksp = x2 a molar
solubility of x = 5.08 × 10−5M (0.00985 g/L). To put that in perspective, it would take a little over 100 L
of water to dissolve 1 g of BaCO3. The cutoff between slightly soluble and insoluble used to establish
1
CRC Handbook of Chemistry and Physics. 87th ed. CRC Press/Taylor and Francis Group, 2006; p 8118.
solubility rules varies somewhat between textbooks, but this would be considered insoluble by any of
them.
Precipitation
When you first encountered the solubility rules, you used them to predict when a precipitation reaction
would occur and wrote net ionic equations for these reactions. Back then you didn’t worry about
concentration; if ions that form an insoluble compound were present, precipitation occurred. Now that
we know about solubility equilibria and saturated/unsaturated/supersaturated solutions, we can dive a
little deeper.
To determine whether a precipitate will form we must calculate the value of Q, the reaction quotient for
the solubility equilibrium. If Q = K the solution is saturated, if Q > K it’s supersaturated, and if Q < K the solution is unsaturated. As we learned in the solutions chapter, a precipitate will only form for a supersaturated solution and will stop forming once the saturation point is reached (Q = K). Example: Consider mixing 25 mL each of aqueous 0.010 M Ba(NO3)2 and 0.010 M Na2CO3. Knowing that BaCO3 is fairly insoluble you might write the following molecular and net ionic equations: Ba(NO3)2 (aq) + Na2CO3 (aq) → BaCO3 (s) + 2 NaNO3 (aq) Ba2+ (aq) + CO32- (aq) → BaCO3 (s) Mixing the solutions dilutes each solution, so [Ba2+] and [CO32-] each decrease to 0.0050 M. Using these values, we can calculate Q. Note the net ionic equation above is the reverse of the Ksp equilibrium. If we’re comparing Q to Ksp , we have to use the expression for Ksp, where the dissociated ions are on the products side. BaCO3 (s) ⇌ Ba2+ (aq) Q = [Ba2+][CO32-]; + CO32– (aq) Q = (0.0050)2 Q = 2.5 × 10−5 ; Ksp = 2.58 × 10−9 Q > Ksp
Because Q > Ksp, the solution is supersaturated and precipitate should form until Q = K. If we
wanted to set know how much solid forms, we could set up an ICE table:
Initial
Change
Equilibrium
BaCO3 (s)
–
⇌
Ba2+ (aq) +
0.0050
-x
0.0050 – x
CO32– (aq)
0.0050
-x
0.0050 – x
2.58 × 10−9 = (0.005 – x)2
Using the quadratic formula, x = 0.00495 M
[Ba2+] = [CO32-] = 0.0050M – 0.00495
[Ba2+] = [CO32-] = 5 × 10−5 M
Initially, [Ba2+] and [CO32-] had the values, so the equilibrium concentrations are also equal and
match the molar solubility of BaCO32-. That’s certainly not always the case. Given the 50 mL of
solution and change of 0.00495M, 0.049 g of solid BaCO3 should form.
0.050 L soln
0.0495 mol Ba2+
L soln
1 mol BaCO3 197.34 g BaCO3
1 mol Ba2+
mol BaCO3
= 0.049 g BaCO3
Common-ion effect
Like other equilibria, the solubility of an ionic compound is subject to LeChâtelier’s principle. For
instance, if Na2CO3 is added to a saturated solution of BaCO3, the equilibrium will shift in response to the
change in [CO32-]. Since Na2CO3 and BaCO3 each form carbonate ions when they dissociate, carbonate is
a common ion and this type of shift in equilibrium as a common ion effect.
In the context of solubility equilibria, the presence of common ions decreases the molar solubility of the
ionic compound. While Ksp is constant at the given temperature, the molar solubility is not. Earlier, we
determined that the molar solubility of BaCO3 was 5.08 × 10−5 M but this assumed the solvent was
pure water. If instead we dissolve BaCO3 in a solution containing 0.10 M Na2CO3 carbonate, our ICE table
looks a bit different:
Initial
Change
Equilibrium
BaCO3 (s)
–
⇌
Ba2+ (aq)
0
+x
x
+
CO32– (aq)
0.10
+x
0.10 + x
𝐾𝑠𝑝 = x (0.10+x)
2.58 × 10−9 = 0.10x + x2
x = 2.58 × 10−8 M
The quadratic formula can be used to solve for x in the calculation above, but when Ksp is small it we can
make the simplifying assumption that x is negligible compared to the soluble source of common ion. In
this case we can assume x