[13%] A sample of solid calcium fluoride is suspended in water in an unreactive container and stirre — Solubility Equilibria and Acid-Base Chemistry Chemistry Question
Problem Context
[13%] A sample of solid calcium fluoride is suspended in water in an unreactive container and stirred until it achieves equilibrium. The pH of the solution is lowered by careful addition of nitric acid, and the pH and concentration of Ca2+(aq) are noted at several points as shown on the graph below. Note that the units on the y axis are millimoles per liter.
Determine the Ksp of CaF2 from the data provided.
Model Answer
From the graph, the molar solubility of CaF2 at high pH is about 0.21 mmol/L, so [Ca2+] = 2.1×10-4 M and [F–] = 2[Ca2+].
Ksp = [Ca2+][F–]2 = (2.1 × 10-4)(4.2 × 10-4)2 = 3.7 × 10-11
Qualitatively, what is the cause for the increase in solubility of CaF2 at low pH?
Model Answer
F– reacts with H+ to form HF, lowering the concentration of F– and shifting the solubility equilibrium to cause more solid CaF2 to dissolve.
From the data provided, determine the Ka of HF.
Model Answer
Choosing any low-pH point should work, but it is better to choose a point with pH < 3 so that the error from reading the plot is minimized. As an example, we can use pH = 2 (1.36 × 10-3 M = [Ca2+]). At this point, [H3O+] = 10–pH = 0.010 M. One can determine the concentration of F– from the Ksp:
Ksp = [Ca2+][F–]2
3.7 × 10-11 = [1.36 × 10-3][F–]2
[F–] = 1.65 × 10-4 M
Since each Ca2+ in solution must be accompanied by two fluorines (in whatever form),
[F–] + [HF] = 2[Ca2+]
[1.65 × 10-4] + [HF] = 2[1.36 × 10-3]
[HF] = 2.56 × 10-3 M
One can then solve for Ka:
Ka = [H3O+][F–] / [HF] = [0.010][1.65×10^-4] / [2.56×10^-3] = 6.4 × 10-4
How many moles of HNO3 must be added to the CaF2/water mixture to achieve a pH = 3.00 in this experiment? The volume of solution is 1.00 L.
Model Answer
At pH = 3, from the graph, [Ca2+] = 3.9 × 10-4 M, so [F–] + [HF] = 7.8 × 10-4 M. At this pH, [H3O+] = 1.0 × 10-3 M, so
6.4 × 10-4 = [0.001][F–] / [HF]
[F–]/[HF] = 0.64
(7.8 × 10-4 – [HF]) = 0.64[HF]
[HF] = 4.8 × 10-4 M
The added nitric acid either protonates fluoride ion or produces H3O+. So
mol added HNO3 = mol HF + mol H3O+
mol added HNO3 = 4.8 × 10-4 + 1.0 × 10-3
Total of 1.5 × 10-3 mol added HNO3
Carbon dioxide dissolves in water at 25 °C and 1 atm pressure to the extent of 0.0345 mol L-1. An aliquot of the solution taken from the above experiment at pH = 5 is stirred under 1 atm CO2 and the pH slowly raised by addition of solid NaOH until CaCO3 just begins to precipitate. What is the pH of the solution at this point? The Ksp of CaCO3 is 8.7 × 10-9, the Ka of aqueous CO2 (“H2CO3”) is 4.3 × 10-7, and the Ka of HCO3– is 4.7 × 10-11.
Model Answer
[Ca2+] = 2.1 × 10-4 M. From the Ksp,
[Ca2+][CO3 2-] = 8.7 × 10-9
[CO3 2-] = 4.1 × 10-5 M
One can relate the concentration of CO3 2- to the concentration of H2CO3 by combining the two acid-base dissociations:
H2CO3 + 2 H2O ⇌ CO3 2- + 2 H3O+
Keq = (Ka of H2CO3)(Ka of HCO3–) = 2.0 × 10-17
[CO3 2-][H3O+]2 / [H2CO3] = 2.0 × 10-17
[4.1 × 10-5][H3O+]2 / [0.0345] = 2.0 × 10-17
[H3O+] = 1.3 × 10-7 M
pH = 6.88