1) What is the difference between solubility and Ksp? 2)How many moles of calciu
ID: 855110 • Letter: 1
Question
1) What is the difference between solubility and Ksp?
2)How many moles of calcium nitrate are in 15.0 mL of 0.200 M calcium nitrate?
3) How many moles of potassium hydroxide are in 15.0 mL of 0.300 M potassium hydroxide?
4) If 15.0 mL of 0.200 M calcium nitrate are mixed with 15.0 mL of 0.300 M potassium hydroxide, how many moles of calcium hydroxide will be expected to be made provided no equilibrium takes place?
5) If only 0.1207 g of calcium hydroxide were collected for the reaction run in question 4, what is the percent yield of the reaction?
Explanation / Answer
1) Solubility describes the concentration of a readily soluble compound. Solubility Product is a quanittative term used to describe the solubility of very sparingly soluble compounds in water. Many compounds are insoluble in water. But there is nothing that is completely insoluble in water. But to describe the solubility in terms of the units used for solubility is cumbersome and of little value. In this case the term "solubility product" is used.
2) Moles of calcium nitrate are in 15.0 mL of 0.200 M calcium nitrate= 0.2 M* (15 ml /1000) = 0.003 moles
3) Moles of potassium hydroxide are in 15.0 mL of 0.300 M potassium hydroxide 0.3 M* (15 ml /1000) = 0.0045 moles
4) The reaction involved : Ca(NO3)2 + 2KOH -------> Ca(OH)2 + 2KNO3
mmoles of Ca(NO3) used = 0.2M *15 ml = 3 mmol
mmoles of KOH = 4.5 mmol
As 1 mmole of Ca(NO3)2 reacts with 2 mmoles of KOH, KOH is the limiting reagent here. Thus, expected mmoles of Ca(OH)2 to be produced = mmoles of KOH used/2 = 4.5/2 = 2.25 mmoles = 0.00225 moles
5) The reaction involved : Ca(NO3)2 + 2KOH -------> Ca(OH)2 + 2KNO3
expected moles of Ca(OH)2 = 0.00225
expected yield of Ca(OH)2 = 0.00225 mol* 74.093 g/mol = 0.1667 g
Thus, percent yield = (0.1207/0.1667)*100 = 72.4 %
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