= 5.16 x 10 3 = 5.2 x 10 3 M. -7 = 1.19 x 10 2 = 1.2 x 10 2 M

Transcription

1 Chemistry B1B Chapter 17 Homework Answers a. The chemical equations are AX A + + X AX 2 A X AX 3 A X b. The concentration of each solution in the compound is 0.25 M. This is because 0.25 mol of the compound was dissolved in 1.0 L of water in each case. Also, when considering the compounds, it is not necessary to consider how each may dissolve in water to form ions. c. In each of these solutions, the concentration of A + ion will be different. Just knowing the stoichiometry of the reaction is not enough. You will need the values of the solubility product constant for each compound. Since the solubility product constant is the same for all the compounds, but each compound dissociates into a different number of ions, the solubilities of the compounds will be different. d. The greatest molar solubility will be for the compound that dissociates into the most ions per formula unit. In this case AX 3 will have the greatest molar solubility. e. The molar solubilities of the compounds are as follows. First, for AX, K sp = [A + ][X ] = s 2 = s = = = = M For AX 2, K sp = [A 2+ ][X ] 2 = 4s 3 = s = = = M For AX 3, K sp = [A 3+ ][X ] 3 = 27s 4 = s = = = M The beaker on the left depicts all the AgCl as individual formula units in solution. This implies AgCl is a soluble nonelectrolye, which is not the case since AgCl as a slightly soluble ionic compound. The center beaker depicts AgCl as a soluble ionic compound that completely dissolves in solution, leaving no AgCl(s). Once again, this cannot be correct since AgCl is slightly soluble. The beaker on the right indicates there are ions of Ag + and Cl present in the solution along with solid AgCl. This is consistent with AgCl being a slightly soluble ionic compound.

2 When a precipitate fails to form when HCl is added to the solution, this indicates that no silver ion is present. When a precipitate fails to form when the solution is acidified and H 2 S is added, this indicates that no copper(ii) ion is present in the solution a. Ca(NO 3 ) 2 is soluble (nitrate salts are soluble). b. AgBr is insoluble (AgBr is an insoluble bromide salt). c. MgI 2 is soluble (iodide salts are generally soluble). d. PbSO 4 is insoluble (PbSO 4 is an insoluble sulfate salt) Assemble the usual concentration table, and substitute from it the equilibrium concentrations into the equilibrium-constant epression. Conc. (M) MgC 2 O 4 (s) Mg C 2 O 4 Starting 0 0 Change Equilibrium K sp = [Mg 2+ ][C 2 O 4 2 ] = ( )( ) = = Let = the molar solubility of PbCrO 4. At the start, before any PbCrO 4 dissolves, the solution contains 0.13 M CrO 2 4. At equilibrium, mol of solid PbCrO 4 dissolves to yield mol Pb 2+ and mol CrO 2 4. Assemble the usual concentration table, and substitute the equilibrium concentrations into the equilibrium-constant epression. As an approimation, assume is negligible compared to 0.13 M CrO 2 4. Conc. (M) PbCrO 4 (s) Pb CrO 4 Starting Change + + Equilibrium [Pb 2+ ][CrO 2 4 ] = K sp ()( ) ()(0.13) = M mol L g 1molPbCrO = = g/l 4 Note that adding to 0.13 M will not change it (to two significant figures), so is negligible compared to 0.13 M.

3 17.46 a. Calculate Q c, the ion product of the solution, using the concentrations in the problem as the concentrations present after miing and assuming no precipitation. Then compare Q c with K sp to determine whether precipitation has occurred. Start by defining the ion product with brackets as used for the definition of K sp ; then use parentheses for the concentrations. Q c = [Sr 2+ ][CO 2 3 ] Q c = (0.012)(0.0015) = Since Q c > K sp ( ), precipitation will occur. b. Calculate Q c, the ion product of the solution, using the concentrations in the problem as the concentrations present after miing and assuming no precipitation. Then compare Q c with K sp to determine whether precipitation has occurred. Start by defining the ion product with brackets as used for the definition of K sp ; then use parentheses for the concentrations. Q c = [Pb 2+ ][Cl ] 2 Q c = (0.0012)(0.041) 2 = Since Q c > K sp ( ), precipitation will occur Because the problem asks you to find [Cl ] at the point when Ag 2 CrO 4 just begins to precipitate, you start with the K sp epression for Ag 2 CrO 4 and ignore temporarily what may have happened to the Cl ion. Also, because the AgNO 3 solution is relatively concentrated, ignore the dilution of the solution of Cl and CrO 2 4 from the addition of AgNO 3. The [Ag + ] just as the Ag 2 CrO 4 begins to precipitate can be calculated from the K sp epression for Ag 2 CrO 4 using [CrO 2 4 ] = M. Thus, for Ag 2 CrO 4, [Ag + ] 2 [CrO 2 4 ] = K sp [Ag + ] 2 [0.015] = [Ag + ] = = M The [Cl ] at this point can be obtained by substituting the [Ag + ] into the K sp epression for AgCl. Therefore, for AgCl, [Ag + ][Cl ] = K sp [ ][Cl ] = [Cl ] = = = M Because this Cl concentration is etremely small compared to the initial M concentration, you can also deduce that essentially all the chloride ion precipitates before the Ag 2 CrO 4 begins to precipitate.

4 Start by calculating the [Ni 2+ ] in equilibrium with the Ni(NH 3 ) 2+ 6 formed from Ni 2+ and NH 3. Then use the [Ni 2+ ] to decide whether or not Ni(OH) 2 will precipitate by calculating the ion product and comparing it with the K sp of for Ni(OH) 2. Assume all the M Ni 2+ reacts with NH 3 to form M Ni(NH 3 ) 2+ 6, and calculate the remaining NH 3. Use these as starting concentrations for the usual concentration table. [0.10 M NH 3 ( M)] = M starting NH 3 Conc. (M) 2+ Ni(NH 3 ) 6 Ni NH 3 Starting Change + +6 Equilibrium Even though this reaction is the opposite of the equation for the formation constant, the formation-constant epression can be used. Simply substitute all the eact equilibrium concentrations into the formation-constant epression; then simplify the eact equation by assuming that is negligible compared to and 6 is negligible compared to K f = [Ni(NH ) ] [Ni ][NH ] Rearrange and solve for : ( ) ( )( ) = 6 (0.0020) 6 ()(0.088) (0.0020) [ (0.088) 6 ] M [Ni 2+ ] Now, calculate the ion product for Ni(OH) 2 : Q c = [Ni 2+ ][OH ] 2 = ( )(0.010) 2 = Because Q c is greater than the K sp of , the solution is supersaturated before equilibrium. At equilibrium, a precipitate will form, and the solution will be saturated The Hg 2+, Ca 2+, and Na + ions can be separated in two steps: (1) Add 0.3 M HCl and H 2 S to precipitate only the HgS, leaving the others in solution. (2) After pouring the solution away from the precipitate, add NH 3 and (NH 4 ) 2 HPO 4 to precipitate only the Ca 3 (PO 4 ) 2 away from the Na + ion, whose phosphate is soluble under these conditions Let equal the molar solubility of MgNH 4 PO 4. Assemble the usual concentration table, and substitute from the table into the equilibrium-constant epression. Conc. (M) MgNH 4 PO 4 (s) Mg 2+ + NH PO 4 Starting Change Equilibrium [Mg 2+ ][NH + 4 ][PO 3 4 ] = K sp ()()() = 3 = = = M a. Molar solubility = = M b mol L g MgNH4PO4 1mol = = g/l

5 Rounded answer: Q c equals (equals K sp, so the solution is saturated). Calculate the concentrations of Mg 2+ and OH. Use a total volume of L L, or L. [Mg 2+ ] = mol L L L = M mol L [OH ] = L = M L Calculate the ion product and compare it to K sp. Q c = [Mg 2+ ][OH ] 2 = ( )( ) 2 = Because Q c equals the K sp of , no precipitation occurs, and the solution is saturated a. Use the solubility information to calculate K sp. The reaction is Pb(IO 3 ) 2 (s) Pb 2+ (aq) + 2IO - 3 (aq) M 2( ) M K sp = [Pb 2+ ][IO 3 ] 2 = [ ][2 ( )] 2 = Set up an equilibrium. The reaction is Pb(IO 3 ) 2 (s) Pb 2+ (aq) 2IO 3 - (aq) + y K sp = [0.15][2y] 2 = Molar solubility = y = = M b. Because of the common-ion effect or a consideration of Le Châtelier s principle, the molar solubility would be less. 2y a. AgBr(s) Ag + (aq) + Br (aq) K sp = Ag + (aq) + 2S 2 O 2 3 (aq) 3 Ag(S 2 O 3 ) 2 K f = AgBr(s) + 2S 2 O 2 3 (aq) Ag(S 2 O 3 ) 3 2 (aq) + Br (aq) K = K sp K f = ( )( ) = 14.5 = 15 mol AgBr = 2.5 g AgBr 1 mol AgBr g AgBr = mol

6 mol Ag(S 2 O 3 ) 3 2 = mol AgBr = mol [Ag(S O ) ][Br ] [S O ] = (0.0133) [S O ] = 14.5 Take the square root of both sides to obtain the moles of S 2 O 3 2 at equilibrium [S O ] = 14.5 [S 2 O 2 3 ] = mol/l Moles of S 2 O 2 3 in Ag(S 2 O 3 ) 3 2 = = mol Total moles of S 2 O 2 3 = = = mol Moles of Na 2 S 2 O 3 added = moles of S 2 O 2 3 = mol After miing, the final volume is 2.00 L. Since the volume doubled, the concentrations are reduced by one-half. Thus, after miing, the initial concentration of Na 2 CO 3 is M, and the concentration of CaCl 2 is M. Calcium chloride precipitates. First, assume that CO 2 3 reacts completely. After the reaction [Ca 2+ ] = M. Now, find the carbonate ion concentration in this solution. The solubility product constant is K sp = = [Ca 2+ ][CO 2 3 ] = ( )() Assume is negligible compared to and solve for. = [CO ] = = = M First, convert the solubilities to molarities using the molar masses. [Na + ] = 3.2 g 1 L + 1 mol Na g = M [NaC 5 H 3 N 4 O 3 ] = 7.0 mg 100 ml 1000 ml 1 L 1 g 1000 mg 1 mol NaC5H3N4O g = M K sp can be calculated from the molar solubility of NaC 5 H 3 N 4 O 3 in aqueous solution. K sp = [Na + ][C 5 H 3 N 4 O 3 ] = ( ) 2 = Now calculate the concentration of urate ion in blood plasma, assuming is negligible compared to M. K sp = = [Na + ][C 5 H 3 N 4 O 3 ] = ( )() = = = M

7 Begin by solving for [H 3 O + ] in the buffer. Ignoring changes in [HC 2 H 3 O 2 ] as a result of ionization in the buffer, you obtain [H 3 O + ] M 0.20 M = M You should verify that this approimation is valid (you obtain the same result from the Henderson-Hasselbalch equation). The equilibrium for the dissolution of MgF 2 in acidic solution is obtained by subtracting twice the acid ionization of HF from the solubility equilibrium of MgF 2. MgF 2 (s) Mg 2+ (aq) + 2F (aq) K sp 2H 3 O + (aq) + 2F (aq) 2HF(aq) + 2 H 2 O(l) 1/(K a ) 2 2H 3 O + (aq) + MgF 2 (s) Mg 2+ (aq) + 2HF(aq) + 2 H 2 O(l) K c = K sp /(K a ) 2 Therefore, K c = ( ) ( ) 2 = In order to solve the equilibrium-constant equation, you require the concentration of HF, which you obtain from the acid-ionization constant for HF. K a = + - [H3O ][F ] [HF] = [F ] [HF] - [F ] [HF] = 17.78, or [F ] = [HF] Let be the solubility of MgF 2 in the buffer. Then [Mg 2+ ] = and [F ] + [HF] = 2. Substituting from the previous equation, you obtain 2 = 17.78[HF] + [HF] = 18.78[HF], or [HF] = 2/18.78 You can now substitute for [H 3 O + ] and [HF] into the equation for K c. K c = [Mg ][HF] (2 /18.78) = [H O ] ( ) = ( ) 3 3 = = = M =