1. Aqueous Reactions and Solution Stoichiometry
CDO High School

2. Solute: the substance that is dissolved in another substance
Solute and solvent
Solute: the substance that is dissolved in another substance
Solvent: the substance that something is dissolved into it

3. Concentrated vs Dilute
Concentration: refers to how much solute is dissolved in the solvent
Concentrated: solutions with a large amount of solute dissolved in solvent
Dilute: solutions with a small amount of solute dissolved the solvent

4. volume of solution in liters
Molarity
Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different.
Molarity is one way to measure the concentration of a solution.
moles of solute
volume of solution in liters
Molarity (M) =

5. Units of Molarity
2.0 M HCl = 2.0 moles HCl 1 L HCl solution 6.0 M HCl = 6.0 moles HCl
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6. Molarity Calculation
NaOH is used to remove potato peels commercially.
If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?
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7. Calculating Molarity
1) 4.0 g NaOH ÷ 40.0 g/mol NaOH = 0.10 mole NaOH 2) 500. mL x 1 L _ = L 1000 mL 3) 0.10 mole NaOH = 0.20 mole NaOH = 0.20 M NaOH L 1 L
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8. Learning Check M1
A KOH solution with a volume of 400 mL contains 2 mole KOH. What is the molarity of the solution?
Drano
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9. Learning Check M2
A glucose solution with a volume of 2.0 L contains 72 g glucose (C6H12O6). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution?
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10. Learning Check M3
Stomach acid is a 0.10 M HCl solution. How many moles of HCl are in 1500 mL of stomach acid solution?
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11. Learning Check M4
How many grams of KCl are present in 2.5 L of M KCl?
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12. Solution M4
3) 2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl 1 L 1 mole KCl
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13. Learning Check M5
How many milliliters of stomach acid, which is 0.10 M HCl, contain 0.15 mole HCl?
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14. Learning Check M6
How many grams of NaOH are required to prepare mL of 3.0 M NaOH solution?
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15. Solution Stoichiometry
The majority of work in research and industry involves solutions. Recall that solutions are easy to handle and are usually easier to control in reactions.
Solution stoichiometry – the procedure for calculating the molar concentration or volume of solution products or reactants

16. Mass A (g) Volume Volume A B (L) (L) Mole Mole A B (mol) (mol)
÷ Molarity A
÷ Molar Mass
X Molar Mass
÷ Molarity B
X Molarity A
X Mole Ratio
X Molarity B
Mole A
(mol)
Mole B
(mol)
X Mole Ratio
÷ Volume B
÷ Volume A
Molarity A
(M)
Molarity B
(M)
X Volume B
X Volume A
X Molar Mass
÷ Molar Mass
Mass B
(g)

17. Example #1
Ammonia and phosphoric acid solutions are used to produce ammonium hydrogen phosphate fertilizer. What volume of 14.8 M NH3(aq) is needed to react with 1000 L of 12.9 M of H3PO4(aq)?
2NH3(aq) H3PO4(aq)  (NH4)2HPO4(aq)
1000 L x mol x 2mol ÷ mol/L = L
1 L mol

18. Example #2
In an experiment, L sample of sulfuric acid solution reacts completely with L of M potassium hydroxide. Calculate the concentration of the sulfuric acid.
H2SO4(aq) KOH(aq)  H2O(l) + K2SO4(aq)
L x mol x 1mol x = mol/L
1 L mol L

19. 2 AgNO3(aq) + K2CrO4(aq)  Ag2CrO4(s) + 2 KNO3(aq)
Example #3
How many grams of silver chromate will form when 120 mL of M silver nitrate are added to potassium chromate?
2 AgNO3(aq) K2CrO4(aq)  Ag2CrO4(s) KNO3(aq)

20. Pb(NO3)2(aq) + 2MgCl2(aq)  PbCl2(s) + Mg(NO3)2(aq)
Example #4
If you mix 200 ml of M Pb(NO3)2 and 300 ml of M MgCl2, how much PbCl2 precipitate will you form?
Pb(NO3)2(aq) + 2MgCl2(aq)  PbCl2(s) + Mg(NO3)2(aq)

21. Move to slide 7 of Molarity dilution ppt