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Molarity, pH, and Stoichiometry of Solutions Chapter 5 part 4.

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Presentation on theme: "Molarity, pH, and Stoichiometry of Solutions Chapter 5 part 4."— Presentation transcript:

1 Molarity, pH, and Stoichiometry of Solutions Chapter 5 part 4

2 Solution Concentration Molarity (M) = mol of solute Liters of solution To make 1.0 M NaCl, dissolve 58.4 g of NaCl in enough water to give a total solution volume 1.0L

3 SHORTCUT! mol M L

4 Problem1 If 25.3 g of Na 2 CO 3 is dissolved in water to make 250. mL of solution. Calculate the molarity of 1) Na 2 CO 3 2) Na + 3) CO 3 2-

5 Problem2 How many moles of CaCl 2 are in 5.00 mL of a 2.00 M CaCl 2 solution? How many moles of Cl - in this volume?

6 Dilution Adding solvent lowers the solution’s concentration M 1 V 1 =M 2 V 2 What vol. of a 0.20 M NaCl is needed to make 500 mL of 0.0010 M NaCl.

7 pH logarithmic scale pH = -log [H + ][ ] = M [H + ] = 10 -pH

8 pH Problems Lemon juice has [H + ]= 0.0032 M, what is the pH? Sea water has a pH of 8.30, what is the concentration of hydrogen ions?

9 Stoichiometry of Aq Rxn What volume of 2.50 M HCl is required to completely react 11.8 g of Zn? Zn + 2 HCl --> ZnCl 2 + H 2

10 Practice 75.0 mL of 0.350 M HCl reacts with an excess of Na 2 CO 3. What volume of CO 2 is produced at STP? 2HCl + Na 2 CO 3 --> 2NaCl + H 2 O+CO 2

11 Titration Chemical analysis used to determine the concentration of a solution or the molar mass of a compound. M a V a = M b V b Moles acid = Moles base

12 Titration (Acid/Base) An indicator, dye that changes color, is used to mark when the reaction is complete. At the equivalence point, there is an equal amount of acid (H + ) and base (OH - ) –pH = 7 w/ strong acids & bases

13 Titration A 0.263 g sample of Na 2 CO 3 requires 28.35 mL of HCl to reach the equivalence point. What is the molarity of HCl? 2HCl + Na 2 CO 3 --> 2NaCl + H 2 O+CO 2


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