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Problem 4.22 (b) Write ionic and net ionic equations for: K 3 PO 4 (aq) + Sr(NO 3 ) 2 (aq) 

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Presentation on theme: "Problem 4.22 (b) Write ionic and net ionic equations for: K 3 PO 4 (aq) + Sr(NO 3 ) 2 (aq) "— Presentation transcript:

1 Problem 4.22 (b) Write ionic and net ionic equations for: K 3 PO 4 (aq) + Sr(NO 3 ) 2 (aq) 

2 Problem 4.34 Balance the following equation and write the corresponding ionic and net ionic equations (a) CH 3 COOH (aq + KOH + 

3 Problem 4.50 (a, b, c, d) Give oxidation numbers for all atoms in: Mg 3 N 2, CsO 2, CaC 2, CO 3 2-

4 Problem 4.54 Predict the outcome of the reactions represent by the following equations by using the activity series, balance the equations. (a) Cu (s) + HCl (aq)  (c ) Mg (s) + CuSO 4 (aq) 

5 Problem 4.56 Classify the following redox reactions. (a) P 4 + 10Cl 2  4PCl 5 (b) 2NO  N 2 + O 2 ( c ) Cl 2 + 2KI  2KCl + I 2

6 SOLUTION STOICHIOMETRY TITRATION SOLUTION CONCENTRATION OF SOLUTION Conc = amount solute/amount solvent DILUTION OF A SOLUTION

7 Solution Stoichiometry The concentration of a solution is the amount of solute present in a given quantity of solvent or solution. M = molarity = moles solute/liters of solution

8 Calculate the moles of KI required to prepare 5.00 X 10 2 mL of a 2.80 M solution.

9 PROBLEM 4.59 Calculate the mass of KI IN GRAMS REQUIRED TO PREPARE 5.00 X 10 2 mL of a 2.80 M solution.

10 10 Preparing a Solution of Known Concentration

11 PROBLEM 4.62 HOW MANY GRAMS OF KOH ARE PRESENT IN 35.0 mL OF A 5.50 M SOLUTION?

12 Problem 4.64 (a) Calculate the molarity of each of the following solutions. (a) 6.57 grams of ethanol (C 2 H 5 OH) in 1.50 x 10 2 ml of solution.

13 13 Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution. Dilution Add Solvent Moles of solute before dilution (i) Moles of solute after dilution (f) = MiViMiVi MfVfMfVf =

14 How would you prepare 60.0 mL of 0.200 M HNO 3 from a stock solution of 4.00 M HNO 3 ? M i V i = M f V f

15 PROBLEM 4.70 WATER IS ADDED TO 25.0 ML OF A 0.866 M KNO3 SOLUTION UNTIL THE VOLUME OF THE SOLUTION IS EXACTLY 500 ML. WHAT IS THE CONCENTRATION OF THE FINAL SOLUTION?

16 16 Titrations In a titration a solution of accurately known concentration is added gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point – the point at which the reaction is complete Indicator – substance that changes color at (or near) the equivalence point Slowly add base to unknown acid UNTIL the indicator changes color

17 17 Titrations can be used in the analysis of Acid-base reactions Redox reactions H 2 SO 4 + 2NaOH 2H 2 O + Na 2 SO 4 5Fe 2+ + MnO 4 - + 8H + Mn 2+ + 5Fe 3+ + 4H 2 O

18 TITRATIONS: EQUIVALENCE POINT MOLES OF ACID = MOLES BASE (MOLES/LITER) ( LITERS) = (MOLES/LITER) (LITERS

19 PROBLEM 4.87 CALCULATE THE VOLUME IN mL OF A 1.420 M NaOH SOLUTION REQUIRED TO TITRATE THE FOLLOWING SOLUTIONS? (a) 25.0 mL OF A 2.430 M HCl SOLUTION (b) 25.00 ML OF A 4.500 M H 2 SO 4 SOLUTION

20 Problem 4.88 (a) What volume of a 0.500 M HCl solution is needed to neutralize each of the following: (a) 10.0 mL of a 2.430 M HCl solution


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