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Chemistry 12 – Sorting out Solubility Problems

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Chemistry 12 – Sorting out Solubility Problems Chemistry 12 – Sorting out Solubility Problems This will help you with the following types of Solubility Problems : 1 -ion concentrations in mixtures (no ppts) 2 -experimental determination of solubility 3 -solubility (s) from Ksp (one compound) 4 -Ksp from solubility (one compound) 5 -predicting precipitates using Trial Ksp 6 -finding maximum concentration of an ion in a solution in which another ion is present 7 -finding minimum concentration of an ion necessary to just start precipitation 8 -finding which precipitate will form first 9 -using titrations to find unknown concentration First-a couple of things to ALWAYS remember: First-a couple of things to ALWAYS remember: 1. Compounds with Nitrate (NO 3) are - ALWAYS soluble (NO 3 is ALWAYS a spectator in unit 3)! First-a couple of things to ALWAYS remember: 1. Compounds with Nitrate (NO 3) are - ALWAYS soluble (NO 3 is ALWAYS a spectator in unit 3)! 2.Compounds with Alkali ions (Na, K, Li etc .) are ALWAYS soluble . (Na +, K + etc. are ALWAYS spectators)! First-a couple of things to ALWAYS remember: 1. Compounds with Nitrate (NO 3) are - ALWAYS soluble (NO 3 is ALWAYS a spectator in unit 3)! 2.Compounds with Alkali ions (Na, K, Li etc .) are ALWAYS soluble . (Na +, K + etc. are ALWAYS spectators)! 3.Always have your solubility and Ksp table handy when doing these problems! 1 -Ion concentrations in mixtures (no ppts ) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH -] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. 1 -Ion concentrations in mixtures (no ppts ) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH -] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! 1 -Ion concentrations in mixtures (no ppts ) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH -] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) 1 -Ion concentrations in mixtures (no ppts ) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH -] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH -] in NaOH 1 -Ion concentrations in mixtures (no ppts ) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH -] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH -] in NaOH -use the Dilution Formula to find the final [Sr(OH)2] 1 -Ion concentrations in mixtures (no ppts ) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH -] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH -] in NaOH -use the Dilution Formula to find the final [Sr(OH)2] - -dissociate the [Sr(OH) 2] to find the [OH ] in [Sr(OH) 2] 1 -Ion concentrations in mixtures (no ppts ) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH -] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH -] in NaOH -use the Dilution Formula to find the final [Sr(OH)2] - -dissociate the [Sr(OH) 2] to find the [OH ] in [Sr(OH) 2] - - -add up the [OH ] from NaOH and the [OH ] from [Sr(OH) 2] to find the final [OH -] . 1 -Ion concentrations in mixtures (no ppts ) In this type there are 1 or 2 compounds and they are BOTH soluble. (There are NO Low solubility Compounds!) An Example: Find the [OH -] when 50.0 mL of 0.20M NaOH is mixed with 250.0 mL of 0.30M Sr(OH) 2. Both of these solutions are soluble (check sol. Table) There is NO compound of Low Solubility involved here. So no equilibrium and no Ksp! In this type of problem: -use the Dilution Formula to find the final [NaOH] in the mixture (final volume is 300.0 mL) -dissociate the NaOH to find the [OH -] in NaOH -use the Dilution Formula to find the final [Sr(OH)2] - -dissociate the [Sr(OH) 2] to find the [OH ] in [Sr(OH) 2] - - -add up the [OH ] from NaOH and the [OH ] from [Sr(OH) 2] to find the final [OH -] . -THE ANSWER IS [OH -] = 0.53 M 2 -Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4 . 2 -Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4 . 1. Subtract # 1 from #2 to find the grams of solid CaSO 4 which dissolved 2 -Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4 . 1. Subtract # 1 from #2 to find the grams of solid CaSO 4 which dissolved 2. Since you are asked for the molar solubility, change these grams to moles. (If you were asked for the solubility in g/L, you could leave them as grams.) 2 -Experimental determination of solubility An example problem is: An experiment was done in which 20.0 mL of saturated CaSO 4 were transferred to a weighed empty evaporating dish and were left to evaporate. The dish and the solid residue were then weighed. The data table is as follows: 1. Mass of empty evaporating dish ……………. 65.340 g 2. Mass of evaporating dish and solid CaSO 4 … 65.363 g 3. Volume of saturated CaSO 4 evaporated …… 20.0 mL Using the data from this experiment, calculate the molar solubility of CaSO 4 . 1. Subtract # 1 from #2 to find the grams of solid CaSO 4 which dissolved 2. Since you are asked for the molar solubility, change these grams to moles. (If you were asked for the solubility in g/L, you could leave them as grams.) 3.
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