Ch.16 - Aqueous Equilibrium WorksheetSee all chapters
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Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: The Ksp of CaSO4 is 4.93 x 10-5. Calculate the solubility (in g/L) of CaSO4(s) in 0.300 M Na2SO4 (aq) at 25 °C.

Solution: The Ksp of CaSO4 is 4.93 x 10-5. Calculate the solubility (in g/L) of CaSO4(s) in 0.300 M Na2SO4 (aq) at 25 °C.

Problem

The Ksp of CaSO4 is 4.93 x 10-5. Calculate the solubility (in g/L) of CaSO4(s) in 0.300 M Na2SO(aq) at 25 °C.

Solution

For this problem, we’re being asked to calculate the solubility (in g/L) of CaSO4(s) in 0.300 M Na2SO4(aq). Since the compounds are ionic compounds, they form ions when dissociating in water. The dissociation of CaSO4 and Na2SO4 in water are as follows:


The sulfate ion, SO42–, has a charge of –2. Calcium is in Group 2A so it’s charge is +2:

CaSO4(s)  Ca2+(aq) + SO42–(aq)


The sulfate ion, SO42–, has a charge of –2. Sodium is in Group 1A so it’s charge is +1:

Na2SO4(s)  2 Na+(aq) + SO42–(aq)


Notice that there is a common ion present, SO42–. The common ion effect states that the solubility of a salt is lower in the presence of a common ion.


We can construct an ICE table for the dissociation of CaSO4. Remember that solids are ignored in the ICE table.



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