On Tuesday, April 3rd, 2012, we worked problem #2 of the Ksp packet. We finished all but part D of the problem. The answer to D is not a numeric value, but rather a conceptual answer. We determined that strontium fluoride compound would begin to precipitate first because of having the lower Ksp value. The percentage of anion (fluoride ion) present when the other precipitate begins to form will be zero. This is because the stronger attraction that strontium ion has for fluoride ion over sulfate ion will cause only strontium flouride precipitate to be formed. Only after the last fluoride ion has been captured and the strontium ion concentration has increased will the precipitate strontium sulfate will be produced.
- Number 3, part B is an ICE problem with an initial concentration of one of the reactants. This will inhibit the right shift that would occur if no products were initially present. "x" is added to both products as a Change because the initial concentration of magnesium ion is zero.
- Number 3, part C gives initial conditions of both ions that can form a precipitate. Whenever initial conditions are given, you must determine the "Q" value to see if it exceeds the Ksp value and a precipitate will form. The molarities and and volumes of each solution are used to determine the millimoles of each ion and then divided by the total volume after the solutions have been poured together. The Q does not exceed Ksp, so no precipitate will form.
- Number 3, part D requires you to think about LeChatelier's Principle. Because the concentration of magnesium decreases after heat is added to the system (the temperature went up), a left shift towards the solid reactant had to occur. Since the equilibrium system must try and remove some of the heat added and the left shift occured, heat must be a product. This makes the dissolving of magnesium fluoride an exothermic reaction.
Number 4, part B (i) asks for the solubility of zinc hydroxide in a solution that already contains hydroxide. The fact that the solution already contains hydroxide ion is that a pH with a value greater than 7 was given. Finding the initial concentration of is shown in the solution to the problem. The presence of hydroxide initially inhibits the right shift of the zinc hydroxide dissolving. This reduces the amount of zinc hydroxide that can dissolve.- Number 4, part B (ii) is a beast. The first thing that must be determined is if a precipitate will form, thus Q must be found. The millimoles of each ion are found and then divided by the total volume of solution once both solutions have been added together to determine the initial concentration of each ion. The Q ends up being larger than the Ksp, so a precipitate forms. The problem then turns into a limiting and excess reactant problem to determine which ion (zinc or hydroxide) is in excess. A BCE table is used. Know that equilibrium will be established again, so the millimoles of the excess hydroxide will be divided by the total volume of solution from adding the two solutions together. If you can do a problem like 4 B (ii), you ROCK!
The solution to part C in #5 is a conceptual question. The question states that solid still remains in the beaker. This means that the solution is still saturated with silver and bromide ions. Therefore, the concentration must be the same value that was calculated in part B because the system is still at equilibrium even though there is a larger volume of solution.- The answer to part D seems like a silly answer (5.0 grams of AgBr in 37,000 liters of water), but it does make sense if you think of how hard it is to dissolve silver bromide with a fairly low Ksp value. The concentration of the 5.0 grams of AgBr in an unknown amount of water must be equal to the concentration determined in part B. After converting to mole of AgBr, I just solved for the unknown volume of solution.
- Part E only asks you what is observed and not to determine the concentration of any ions after equilibrium is established again. This means that "Q" must be found using the conditions given. The millimoles of silver ion (from silver nitrate) and bromide ion (from sodium bromide) were found by muliplying the volume of each solution by the molarity of each solution and then dividing by the total volume (12.0 mL) after the two solutions are added together. In this problem, the Q exceeds Ksp, thus a precipitate is formed.
- I will do a demonstration on Friday, April 6th to illustrate part F. I will explain it more then.
