3. Match each of the following descriptions with one of the beakers in Model 1. In each case, assume the change in volume as the solid(s) are added is minimal. a. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of water in beaker . b. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.500 M sodium hydroxide solution in beaker . c. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.200 M sodium hydroxide solution in beaker . d. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.200 M calcium nitrate solution in beaker . 4. Based on the solubility product constant, Ksp, for calcium hydroxide given in Model 1, do you expect most of the 1.00 mole sample of solid to dissolve in any of the four beakers

Explanation:

When calcium hydroxide is dissolved in water , it ionizes as follows .

Ca( OH)₂ = Ca⁺² + 2 OH ⁻

When it is dissolved in water which contains minimal OH⁻ , so there is almost no common ion effect . Hence calcium hydroxide is fully dissolved in pure water solvent .

When 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.500 M sodium hydroxide solution in beaker , it is not fully dissolved due to common ion of hydroxide ion ( OH⁻ )

NaOH = Na⁺ + OH⁻

OH⁻ ion from NaOH , suppresses the dissolution of calcium hydroxide .

Similarly

When A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.200 M sodium hydroxide solution in beaker , it is not fully dissolved due to common ion of hydroxide ion ( OH⁻ )  

NaOH = Na⁺ + OH⁻

OH⁻ ion from NaOH , suppresses the dissolution of calcium hydroxide

When 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.200 M calcium nitrate solution , it is not fully dissolved due to common ion of calcium  ion ( Ca⁺² )

Ca( NO₃)₂ = Ca⁺² + 2NO₃⁻

Ca⁺²  ion from Ca( NO₃)₂ , suppresses the dissolution of calcium hydroxide .

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8