Question 1 (c)

(c) A 50.0 mL sample of distilled water is added to the solution described in part (b), which is in a beaker with some solid AgBr at the bottom. The solution is stirred and equilibrium is reestablished. Some solid AgBr remains in the beaker. Is the value of \[Ag+\] greater than, less than, or equal to the value you calculated in part (b) ? Justify your answer. The value of \[Ag+\] after addition of distilled water is equal to the value in part (b). The concentration of ions in solution in equilibrium with a solid does not depend on the volume of the solution. One point is earned for the correct answer with justification.

Question 1 (e)

(e) A student mixes 10.0 mL of 1.5 x 10-4 M AgN03 with 2.0 mL of 5.0 x 10-4 M NaBr and stirs the resulting mixture. What will the student observe? Justify your answer with calculations. (10.0 x 10-4 M) = 1.3 M \[Ag+\] = 12.0mL (2.0 x 104M) = 8.3 x 10-5 M 12.0mL Q = = (1.3 x 10-4 x 10-5M) = 1.1 x 10-8 -13 . a precipitate will form. One point is earned for calculation of concentration of ions. One point is earned for calculation of Q and conclusion based on comparison between Q and KS One point is earned for indicating the precipitation of AgBr.

17.61 Precipitation and Separation of Ions Equilibrium can be achieved starting with the substances on either side of a chemical equation. For example, the equilibrium that exists between BaS04(s), Ba2+ (aq), and S042— (aq) (Equation 17.15), can be achieved either by starting with BaS04(s) or by starting with solutions containing Ba2+ and S042— . If we mix, say, a BaC12 aqueous solution with a NalS04 aqueous solution, BaS04 might precipitate out. How can we predict whether a precipitate will form under various conditions? Recall that we used the reaction quotient Q in Section 15.6 to determine the direc- tion in which a reaction must proceed to reach equilibrium. The form of Q is the same as the equilibrium-constant expression for a reaction, but instead of only equilibrium concentrations, you can use whatever concentrations are being considered. The direc- tion in which a reaction proceeds to reach equilibrium depends on the relationship between Q and K for the reaction. If Q < K, the product concentrations are too low and reactant concentrations are too high relative to the equilibrium concentrations, and so the reaction will proceed to the right (toward products) to achieve equilibrium. If Q > K, product concentrations are too high and reactant concentrations are too low, and so the reaction will proceed to the left to achieve equilibrium. If Q = K, the reac- tion is at equilibrium. For solubility-product equilibria, the relationship between Q and Ksp is exactly like that for other equilibria. For K sp reactions, products are always the soluble ions, and the reactant is always the solid. Therefore, for solubility equilibria, • If Q = K , the system is at equilibrium, which means the solution is saturated; this is the highest concentration the solution can have without precipitating. • If Q < K the reaction will proceed to the right, towards the soluble ions; no precipitate will form. If Q > Ksp, the reaction will proceed to the left, towards the solid; precipitate will form. For the case of the barium sulfate solution, then we would calculate Q = Ba2+ lS042 and compare this quantity to the K sp for barium suhte.

SAMPLE EXERCISE 17.16 Predicting Whether a Precipitate Forms Does a precipitate form when 0.10 L of 8.0 X 10—3 M is added to 0.40 L of 5.0 X 10—3 M Na7SOd? SOLUTION Analyze The problem asks us to determine whether a precipitate forms when two salt solutions are combined. Plan We should determine the concentrations of all ions just after the solutions are mixed and compare the value of Q with Ksp for any potentially insoluble product. The possible metathesis products are Solve When the two solutions are mixed, thev01umeisO.10L + 0.40L = 0.50L. The number of moles of Pb2+ in 0.10 L of 8.0 X The concentration of PV in the 0.50-L 8.0 X lo- (0.10 L) mol mixture is therefore The number of moles of SO? — 5.0 X 10-3MNaS04is in 0.40 L of 8.0 X 10-4m01 Pb2+ = 0.50 L 5.0 X 10-3m01 (0.40 L) PbS04 and NaN03. Like all sodium salts NaN03 is soluble, but PbS04 has a K of 6.3 X 10—7 (Appendix D) and will precipitate if the Pb2+ and SOd — concentrations are high enough for Q to exceed K = 8.0 X 10-4m01 = 1.6 X lo- M mol — 2.0 X lo- Therefore Because Q > K 2.0 X 10-3m01 S04 2 = 4.0 X 10-3M 0.50 L Q = = (1.6 X X 10-3) = 6.4 X 10-6 PbS04 precipitates.

Question 5 (f)

(D Ethanol is completely soluble in water, whereas ethanethiol has limited solubility in water. Account for the difference in solubilities between the two compounds in terms of intermolecular forces. Ethanol is able to form strong hydrogen bonds with water whereas ethanethiol does not have similar capability. The formation of hydrogen bonds increases the attraction between molecules of ethanol and molecules of water, making them more soluble in each other. Note: The answer must clearly focus on the solute- solvent interaction. Just the mention of hydrogen bonding does not earn the point. One point is earned for the correct explanation.

Influence of substituents on the solubility • Polar groups such as —OH capable of hydrogen bonding with water molecules impart high solubility • Non-polar groups such as —CH3 and —Cl are hydrophobic and impart low solubility. • Ionization of the substituent increases solubility, e.g. —COOH and —NH2 are slightly hydrophilic whereas —COO— and —NH3 are very hydrophilic. -OCH2CHs -OCHg XHO vCOOH -OH tion Hydrophobi€ H yd rophobic Hydrophobic Slightly hydrophilic Slightly hydrophilic Hydrophilic Slightly hydrophilic Very h yd a..philic Hydrophilic Very Very hydrophilic

Question 6 (b)

(b) Which species, Zn or Zn2+ , has the greater ionization energy? Justify your answer. Zn2+ has the greater ionization energy. The electron being removed from Zn2+ experiences a larger effective nuclear charge than the electron being removed from Zn because Zn2+ has two fewer electrons shielding the nucleus. OR It takes more energy to remove a negatively charged electron from a positive ion than from a neutral atom. One point is earned for identifying Zn2+ with justification.

Question 6 (g)

(g) If the concentration of in the Al(s)/A13+(aq) half-cell is lowered from 10M to 0.01 M at 250C, does the cell voltage increase, decrease, or remain the same? Justify your answer. Lowering \[A13+\] causes an increase in the cell voltage. The value of O will fall below 1.0 and the 102 term in the Nernst equation will become negative. This causes the value of Ecell to become more positive. OR A decrease in a product concentration will increase the spontaneity of the reaction, increasing the value of E cell • One point is earned for indicating that E cell increases. One point is earned for the correct justification.

results matching ""

    No results matching ""