Titration Example Problem Suppose that 10.0g of an unknown monoprotic weak acid, HA, is dissolved in 100 mL of water. To reach the equivalence point, 100.0mL of 0.10 M NaOH was used. After the addition of 50.0 mL, the pH of the solution was found to be 4.00.

Titration Example Problem Suppose that 10.0g of an unknown monoprotic weak acid, HA, is dissolved in 100 mL of water. To reach the equivalence point, 100.0mL of 0.10 M NaOH was used. After the addition of 50.0 mL, the pH of the solution was found to be 4.00.

Titration Example Problem Suppose that 10.0g of an unknown monoprotic weak acid, HA, is dissolved in 100 mL of water. To reach the equivalence point, 100.0mL of 0.10 M NaOH was used. After the addition of 50.0 mL, the pH of the solution was found to be 4.00.

Molar Solubility- The amount of a compound that is soluble. Mols/L Calculated from K sp

Common Ion Effect and Solubility and Kf Just like in K eq, if you add it to a salt solution with a common ion it changes the solubility Lowered Solubility Fill in concentration to K sp equation You may make an ice chart if it helps, but it is not necessary. Increased Solubility: Kf If a solution’s ion reacts with an ion from the insoluble compound the reactions can be combined by adding reactions, and multiplying the two K together. Increased Solubility: Acid Conceptual only Le Chatlier’s Reminder:

Common Ion vs. Increasing solubility. Le Chatelier’s Principle. How would you decrease the solubility of AgCl? How many grams of AgCl dissolve in 0.5L of 0.1M NaCl? Add a common ion: either Ag + or Cl - 0.1M 0M +x Ksp AgCl=5x x-x 0.1+xx

Conceptual Topics: Common Ion vs. Increasing solubility. Le Chatelier’s Principle. How would you decrease the solubility of magnesium carbonate? How would you increase the solubility? Decrease: add a common ion Increase: add something that “removes” some product. What reacts with CO 3 - ? An Acid!

Common Ion Effect and Solubility You can have two different salt solutions that are perfectly soluble, but when mixed form a precipitate See the picture for how this happens You can calculate if this will happen Q<K no precipitate Q=K saturated solution Q>K precipitate We can see if you mix two salt solutions if a precipitate will form. This is like using Q to determine which way the reaction will go.

Ksp AgCl=5x10 -13

Balance the following redox reaction in an acidic solution. Decide which is the oxidizing agent and which is the reducing agent. Describe the flow of electrons. Is the reaction spontaneous? Find G and K. ZnS(s) + NO 3 - (aq)  Zn 2+ (aq)+ S (s)+ NO (g) Balance all except O and H: Done! Use oxidation numbers to balance electrons Add H 2 O to balance O Add H+ to balance Hs 2e*1=2 3e*1=3 ( )*3=6 ( )*2= H 2 O 8H + + Oxidized: S Reducing Agent: ZnS Reduced: N Oxidizing agent: NO 3 - Electrons are given up by S 2- to the N

Balance the following redox reaction in an acidic solution. Decide which is the oxidizing agent and which is the reducing agent. Describe the flow of electrons. Is the reaction spontaneous? Find G and K. ZnS(s) + NO 3 - (aq)  Zn 2+ (aq)+ S (s)+ NO (g) H 2 O 8H + + From Reduction Table: S/S 2- = NO 3 - /NO = E= = V

Cells Decide which is the oxidizing agent and which is the reducing agent for the following galvanic cell. Is the reaction spontaneous? What is the Ecell? What would happen to the mass of the anode vs the mass of the electrodes at the cathode. Describe the electron flow. Pb 2+ /Pb = -0.13V Au 1+ /Au = V  Higher E red =cathode Galvanic=spontaneous: so Yes spontaneous! Au 1+ is reduced: making it the oxidizing agent Pb is oxidized: making it the reducing agent E cell = = 1.82 V Au electrode mass increases: Pb electrode mass decreases Electrons flow from Pb to Au