UNIT 5 Aqueous Reactions and Solution Stoichiometry Molarity.

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Presentation transcript:

UNIT 5 Aqueous Reactions and Solution Stoichiometry Molarity

Concentration of an Aqueous Solution - Molarity An important property of a solution is its concentration : the amount of solute dissolved in a given quantity of solvent. There are many ways to express concentration. We will learn one: molarity. Molarity (symbol M) = moles of solute volume of solution in liters Molarity can be treated as a conversion factor (like density) to go from volume of a solution to moles of solute and vice versa.

Two Ways Calculate Moles You now know two ways to calculate the number of moles of a substance. 1. If you have a known mass of the substance, you divide by the molar mass to get moles. mass of A x 1 mol A = mol A molar mass of A 2. If you have a known volume of a solution of known molarity, you multiply the volume of the solution in liters by the molarity to get moles. liters of solution of A x molarity of A = mol A

Calculations Involving Molarity How to prepare 1.00 liter of a 1.00 M solution of copper(II) sulfate: 1.00 L x 1.00 mole CuSO 4 = 1.00 mole CuSO 4 1 L solution 1.00 mole CuSO 4 x g CuSO 4 = 160. g CuSO 4 1 mole CuSO 4 Put 160. g CuSO 4 in a 1L volumetric flask. Fill about ¼ full with deionized (DI) water and swirl to dissolve. Fill to the mark with DI water and up end the flask to mix.

Calculations Involving Molarity How to prepare mL of a 1.00 M solution of copper(II) sulfate: mL x 1 L x 1.00 mole CuSO 4 = mole CuSO mL 1 L solution mole CuSO 4 x g CuSO 4 = 39.9 g CuSO 4 1 mole CuSO 4 Put 39.9 g CuSO 4 in a 250 mL volumetric flask and follow previous procedure. Both flasks contain 1.00M CuSO 4.

Calculations Involving Molarity What is the molarity of a solution made by dissolving 25.0 g of NaCl in enough water to make 125 mL? molarity = moles NaCl = 25.0 g NaCl x 1 mole NaCl vol soln in L g NaCl 125 mL x __1L___ 1000 mL = moles NaCl L = 3.42 M

Molarity of Electrolytes Molarity is just a unit that describes concentration. It can describe the concentration of the entire compound or any part of that compound. A M solution of potassium carbonate has what concentration of potassium ions? Carbonate ions? M K 2 CO 3 = mol K 2 CO 3 x 2 mol K + = 1.00 M K + 1 L soln 1 mol K 2 CO M K 2 CO 3 = mol K 2 CO 3 x 1 mol CO 3 2- = M CO L soln 1 mol K 2 CO 3

Using Molarity to Convert Between Moles and Volume We have learned how to use the stoichiometry of a reaction to predict the yield of a product in grams or moles. We can now extend this knowledge to include reactions involving solutions as reactants.

How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO 3 ? The steps are the same as with the stoichiometry we learned in Chapter Write and balance the equation for the reaction: HNO 3 (aq) + KOH (aq)  H 2 O (l) + KNO 3 (aq) Using Molarity to Convert Between Moles and Volume

How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO 3 ? 2. Write what you know and what you need to know: 15.0 mL 6.0M ? mL 1.50M HNO 3 (aq) + KOH (aq)  H 2 O (l) + KNO 3 (aq) 3. Convert “what you know” to moles mL 1 L 6.0 mol HNO mL 1 L solution Using Molarity to Convert Between Moles and Volume

How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO 3 ? 15.0 mL 6.0M ? mL 1.50M HNO 3 (aq) + KOH (aq)  H 2 O (l) + KNO 3 (aq) 4. Convert to “moles of what you need to know” using the stoichiometry of the reaction: 15.0 mL 1 L 6.0 mol HNO 3 1 mol KOH 1000 mL 1 L solution 1 mol HNO 3 Using Molarity to Convert Between Moles and Volume

How many mLs of 1.50 M KOH are required to neutralize 15.0 mL of 6.0 M HNO 3 ? 15.0 mL 6.0M ? mL 1.50M HNO 3 (aq) + KOH (aq)  H 2 O (l) + KNO 3 (aq) 5. Convert your answer from moles to the required units. We have been asked for volume. We have moles. Molarity lets us convert between the two: 15.0 mL 1 L 6.0 mol HNO 3 1 mol KOH L 1000 mL = 60. mL 1000 mL 1 L 1 mol HNO mol KOH 1 L 60. mL of 1.50M KOH are required. Using Molarity to Convert Between Moles and Volume

Suppose the question had been “How many grams KOH are required to neutralize 15.0 mL of 6.0 M HNO 3 ?” Only step 2 and step 5 change! 15.0 mL 6.0M ? g KOH (step 2) HNO 3 (aq) + KOH (aq)  H 2 O (l) + KNO 3 (aq) 5. Convert your answer from moles to the required units mL 1 L 6.0 mol HNO 3 1 mol KOH g KOH = 5.1 g KOH 1000 mL 1 L 1 mol HNO 3 1 mol KOH Using Molarity to Convert Between Moles and Volume

Dilutions liters of solution of A x molarity of A = mol A This equality is helpful in dealing with dilutions. Often we are called upon to make a solution of a known concentration not from a dry chemical, but from another solution. How do we make mL of 2.50M HNO 3 from concentrated (12.0 M) HNO 3 ?

200.0 mL x 1L x 2.50 mol HNO 3 = mol HNO mL1L We must determine how many mLs of 12.0M HNO 3 contains this same number of moles mol HNO 3 x_____1L_____ = L 12.0 mol HNO 3 Answer: 41.7 mL Dilutions

How do we make mL of 2.50M HNO 3 from concentrated (12.0 M) HNO 3 ? The answer is to take 41.7 mL of 12.0 M HNO 3 and add enough water to make mL of solution…actually, you would add the acid to the water! Dilutions

Dilutions – A Quicker Way How do we make mL of 2.50M HNO 3 from concentrated (12.0 M) HNO 3 ? If we set this up algebraically, we get 2.50M x L = 12.0M x ?L. With the answer included, we have 2.50M x L = 12.0M x L.

Dilutions – A Quicker Way The general form for dilutions is M 1 V 1 = M 2 V 2 where V = volume (mL or L, just keep them the same) and M = molarity (NOT MOLES!!!)

Dilutions – Examples M 1 V 1 = M 2 V 2 The Dilution Equation How many mLs of 0.750M HCl are needed to make mL of 0.125M HCl? How much water should be used? (0.125M)(150.0mL) = (0.750M) (x mL) x = 25.0 mL Water used = – 25.0 = mL

M 1 V 1 = M 2 V 2 What is the concentration of a solution made by diluting mL of 18M sulfuric acid to mL? (18M)(250.0 mL) = (x M)( mL) x = 3.0M Dilutions – Examples

Dilutions – Mixing Two Solutions Often reactions are carried out by mixing two solutions (one for each reactant) together. It then becomes necessary to calculate the concentration of the reacting species in the resulting solution. Example: mL of M ammonium phosphate are mixed with mL of M sodium phosphate. What is the concentration of each ion in the resulting solution? (NH 4 ) 3 PO 4 (aq) + Na 3 PO 4 (aq)  NR (no reaction…why?) This is just a dilution!

Dilutions – Mixing Two Solutions Example: mL of M ammonium phosphate are mixed with mL of M sodium phosphate. What is the concentration of each ion in the resulting solution? molarity = ________moles ion____________ volume of the final solution in liters volume of the final solution = = mL

Example: mL of M ammonium phosphate are mixed with mL of M sodium phosphate. What is the concentration of each ion in the resulting solution? final volume = mL = L molarity = moles ion L Let’s use the dilution equation to find the concentration of ammonium phosphate (NH 4 ) 3 PO 4 in the resulting solution: mL (0.820 M) = mL M 2 M 2 = mol (NH 4 ) 3 PO 4 per liter of the final solution Dilutions – Mixing Two Solutions

Example: mL of M ammonium phosphate are mixed with mL of M sodium phosphate. What is the concentration of each ion in the resulting solution? final volume = mL = L Let’s use the dilution equation to find the concentration of sodium phosphate Na 3 PO 4 in the resulting solution: mL (0.750 M) = mL M 2 M 2 = mol Na 3 PO 4 per liter of the final solution Dilutions – Mixing Two Solutions

Example: mL of M ammonium phosphate are mixed with mL of M sodium phosphate. What is the concentration of each ion in the resulting solution? Find the concentration of NH 4 + in the final solution: mol (NH 4 ) 3 PO 4 x 3 mol NH 4 + = 1.81 mol NH 4 + L of final soln 1 mol (NH 4 ) 3 PO 4 L 1.81 mol NH 4 + = 1.81 M NH 4 + L Dilutions – Mixing Two Solutions

Example: mL of M ammonium phosphate are mixed with mL of M sodium phosphate. What is the concentration of each ion in the resulting solution? Find the concentration of Na + in the final solution: mol Na 3 PO 4 x 3 mol Na + = mol Na + L of final soln 1 mol Na 3 PO 4 L mol Na + = M Na + L Dilutions – Mixing Two Solutions

Example: mL of M ammonium phosphate are mixed with mL of M sodium phosphate. What is the concentration of each ion in the resulting solution? The phosphate comes from both solutions. To find its concentration in the final solution, add the concentrations of (NH 4 ) 3 PO 4 and Na 3 PO 4 in the final solution (why?): M of PO 4 3- in final solution = M of (NH 4 ) 3 PO 4 + M of Na 3 PO 4 = = M PO 4 3- in final solution Dilutions – Mixing Two Solutions Both molarities are for the final solution.

Titration Titration is a way to determine the concentration of a solution by reacting it with a known amount of a chemical (the standard) and monitoring the point of stoichiometric equivalence with an indicator. Titration can be performed on any solution that gives a product which can be monitored: acid-base titrations redox titrations precipitation titrations

Titration Titration is a way to determine the concentration of a solution by reacting it with a known amount of a chemical (the standard) and monitoring the point of stoichiometric equivalence with an indicator. You have already performed the calculations for an acid-base titration. This is just a stoichiometric problem with the concentration of one of the reactants known.

Titration – Example 1 KHP (potassium acid phthalate) is a primary standard with a molar mass of g. NaOH is a secondary standard, the concentration of which is found by titration with a known amount of KHP, according to the following equation: mL ? M g NaOH (aq) + KHP (s)  KNaP (aq) + H 2 O(l) If mL of NaOH solution are needed to reach a phenolphthalein endpoint in the titration of g of KHP, what is the molarity of the NaOH solution?

Titration – Example mL ? M g NaOH (aq) + KHP (s)  KNaP (aq) + H 2 O(l) ? M = mol NaOH volume of NaOH solution in L g KHP 1 mol KHP 1 mol NaOH _______ 1000 mL = mol NaOH g KHP 1 mol KHP mL 1 L L of solution This is the same as M NaOH.

Titration – Example 2 The NaOH solution we just titrated is a secondary standard and can be used to titrate acid samples such as vinegar, a solution of acetic acid in water. If mL of the vinegar is neutralized by mL of the NaOH solution, what is the molarity of the acetic acid in the vinegar? mL M mL ? M NaOH (aq) + HC 2 H 3 O 2 (aq)  NaC 2 H 3 O 2 (aq) + H 2 O(l)

Titration – Example mL M mL ? M NaOH (aq) + HC 2 H 3 O 2 (aq)  NaC 2 H 3 O 2 (aq) + H 2 O(l) The key to most titration problems is to keep track of which solution you need at what time. molarity of acetic acid = moles acetic acid volume of acetic acid (vinegar) in L mL NaOH 1 L mol NaOH 1 mol acetic acid ________ 1000 mL = 1000 mL 1L NaOH soln 1 mol NaOH mL 1 L mol acetic acid L This is the same as M acetic acid.

Titration – Example 3: Polyprotic Acids If mL of a solution of malonic acid (H 2 C 3 H 2 O 4 ) is neutralized by mL of the NaOH solution, what is the molarity of the malonic acid? Note the position of the acid H’s in malonic acid mL M mL ? M 2NaOH (aq) + H 2 C 3 H 2 O 4 (aq)  Na 2 C 3 H 2 O 4 (aq) + 2H 2 O(l) mL NaOH 1 L mol NaOH 1 mol H 2 C 3 H 2 O 4 _________ 1000 mL = 1000 mL 1L NaOH soln 2 mol NaOH mL 1 L mol H 2 C 3 H 2 O 4 L This is the same as M H 2 C 3 H 2 O 4.