Chapter 18 Homework Answers

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Dwight Johnson
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1 Chapter 18 Homework Answers (a) AgI(s) Ag + + I K sp = [Ag + ][I ] (b) Ag PO (s) Ag + + PO K sp = [Ag + ] [PO ] (c) PbCrO (s) Pb + + CrO K sp = [Pb + ][CrO ] (d) Al(OH) (s) Al + + OH K sp = [Al + ][OH ] (e) ZnCO (s) Zn + + CO K sp = [Zn + ][CO ] (f) Zn(OH) (s) Zn + + OH K sp = [Zn + ][OH ] 18.5 To solve this problem we need to realize that the concentration of the solution is equal to the number of moles of solid recovered divided by the volume of the solution, i.e., 0.69 g CaCrO 1 mole CaCrO 1000 ml [ CaCrO ] = = M 156 ml g CaCrO 1 L The equilibrium for this problem is CaCrO (s) Ca + + CrO K sp = [Ca + ][CrO ] = (0.067) = Ag CrO (s) Ag + + CrO [Ag + ] [CrO ] I C + x + x E + x + x K sp = (x) (x) = x = , x = = M 18.7 First determine the molar solubility of the M X salt. M X (s) M + (aq) + X (aq) K sp = [M + ] [X ] [M + ] [X ] I C + x + x E + x + x K sp = (x) (x) = = 108x 5 x = M The molar solubility of this compound is moles/l We want the molar solubility of the M X compound to be twice the value just calculated or moles/l. We need to solve the equilibrium expression: M X (s) M + (aq) + X (aq) K sp = [M + ] [X ] [M + ] [X ] I C + x + x E + x + x K sp = (x) (x) = x So, K sp = ( ) = where x = M

2 18. AuCl (s) Au + + Cl K sp = [Au + ][Cl ] = (a) (b) (c) (d) let x = [Au + ]; then [Cl ] = x K sp = (x)(x) = 7x 5 x =. 10 = M 7 The molar solubility of AuCl is M in H O. [Au + ] = x; [Cl ] = x K sp = (x)( x) : Assume x << K sp = (x)(0.010) x = M The molar solubility of AuCl is M in M HCl. [Au + ] = x; [Cl ] = x K sp = (x)( x) : Assume x << 0.00 K sp = (x)(0.00) x = M The molar solubility of AuCl is M in M MgCl. [Au + ] = x; [Cl ] = x K sp = ( x)(x) : Assume x << K sp = (0.010)(x) 5 x =. 10 = M 0.7 The molar solubility of AuCl is M in M Au(NO ) (a) Ca(OH) (s) Ca + (aq) + OH (aq) K sp = [Ca + ][OH ] = let x = [OH ], [Ca + ] = x K sp = ( x)(x) = By successive approximations, x = The molar solubility of Ca(OH) in 0.10 M CaCl is moles/l. (b) Ca(OH) (s) Ca + (aq) + OH (aq) K sp = [Ca + ][OH ] = let x = [Ca + ], [OH ] = x K sp = (x)(x ) = By successive approximations, x =.9 x 10 The molar solubility of Ca(OH) in 0.10 M NaOH is.9 x 10 moles/l This problem is similar to except that the K sp constants are closer in value. We first determine the minimum amount of SO that must be added to initiate the precipitation of CaSO. CaSO will precipitate after SrSO due to its larger value for K sp : K sp (CaSO ) = and K sp (SrSO ) = (see Table 18.1) + (a) Let x = Ca ; + K sp = Ca SO = (0.15)(x) = When the SO x = SO =. 10 M =. 10 the CaSO will start to precipitate. Now we ask, what is the Sr + if SO =. 10 M?

3 SrSO (s) Sr + (aq) + SO (aq) + K sp = Sr SO [Sr + ] [SO ] I. 10 C + x x E + x x + K sp = Sr SO = (x)( x) = For this problem, we must solve the quadratic equation and we determine that x =. 10 M. Thus, the [Sr + ] =. 10 M when the CaSO starts to precipitate. (b) Initially the solution had a concentration of 0.15 M. The solution now has a [Sr + ] =. 10 M. So, the percentage of Sr + precipitated is; % = 99.7 % (a) Ag + (aq) + S O (aq) Ag(S O ) (aq) Ag(S O ) form + Ag SO (b) (c) Zn + (aq) + NH (aq) Zn(NH ) + (aq) Sn + (aq) + S (aq) SnS (aq) Zn(NH ) + form + Zn NH SnS form + Sn S The applicable equilibria are as follows: AgI(s) Ag + (aq) + I + 17 (aq) K sp = Ag I = Ag ( I) Ag + (aq) + I (aq) AgI 11 (aq) K form = = Ag I The two equations above may be combined and K c found as follows: Ag ( I) AgI(s) + I (aq) Ag(I) (aq) K c = = K sp K form = I If all of the AgI dissolves, it will be in the form of Ag(I), therefore the concentration of Ag(I) is: 0.00 mol Ag ( I) [Ag(I) ] = = 0.00 M Ag(I) L solution [CN 0.00 ] = =. 10 M

4 The amount of KI that must be added is: (. 10 M)(0.100 L) = 00 mol KI g KI = (00 mol KI)(166 g/mol) = g KI The student would not collect PbCl by adding HCl to the final solution. The Pb + would be preciptated in step (1) when H S was added to the acidified solution. This precipitate was separated from the remaining solution. The first precipitate would consist of PbS and CdS while the second precipitate consists of NiS Initially, both Ag + and HC H O are at 1.0 M concentrations. These values will be used to determine if the AgC H O will precipitate. First, determine the concentration of the acetate ion from the equilibrium: HC H O H + + C H O + H CHO K a = = HCHO [HC H O ] [H O + ] [C H O ] I 1.0 C x + x + x E 1.0 x x x Assume x << 1.0 [x][x] 5 - K a = = x =. 10 = [CHO ] 1.0 [ ] Next, using the concentration of the acetate ion, determine whether or not a precipitate will form. AgC H O (s) Ag + + C H O K sp = [Ag + ][C H O ] Q = [Ag + ][C H O ] before equilibrium is established Q = (1.0 M)(. 10 M) =. 10 Q > K sp therefore a precipitate will form To answer this question we can use cation groupings from qualitative analysis. Ag + and Pb + are group I ions. These ions can be separated from the others by adding Cl to the solution. AgCl and PbCl will precipitate and can be filtered off from the solution. AgCl can be separated from PbCl by adding 6 M NH. AgCl will react forming Ag(NH ) + while PbCl will remain a solid. Cu + and Bi + can be separated by adding 0.1 M H S and 6M HCl. This will precipitate CuS and Bi S which can be filtered The solids can be converted to soluble species by adding HNO. Then, base can be added to precipitate hydroxides; Cu(OH) and Bi(OH). The addition of 6M NH will convert the Cu(OH) to Cu(NH ) + leaving the solid Bi(OH). Then add base to the remaining solution along with H S to precipitate CoS and MnS. Filter the solution and dissolve the solid be adding acid. Cobalt can be separated by adding KNO which forms a yellow precipitate while Mn + remains in solution. Finally, Ba + and Ca + can be separated by adding acid to make a weakly acidic solution and then adding Na CrO. BaCrO will precipitate while Ca + will remain in solution.

5 The volume of a cone is given by V=(1/)π r h 1 1 in.5 cm V = x.115 x ( 5 ft ) x 16ft x x ft in V = 1.19 x 10 7 cm The mass of the cone is: 1.19 x 10 7 cm x.71 g cm - =. x 10 7 g CaCO Saturated CaCO has a molar concentration of Ca + equal to: [Ca + ] = [CO - ] = K 1/ sp = (. x 10-9 ) 1/ = 5.8 x 10-5 M Mass of CaCO dissolved in a liter of saturated solution is: 5.8 x 10-5 mol L -1 x ( g/ mol CaCO ) = 5.8 x 10 - g L -1 The volume of solution required to build the cone would be equal to:. x 10 7 g CaCO x (1 L/5.8 x 10 - g) = 5.6 x 10 9 L 10 drops have a volume of 1 ml. (10 drops/1 ml) x (1000 ml/l) = drops/l The number of drops required to deposit enough CaCO is given by: (10000 drops/l) x 5.6 x 10 9 L = 5.6 x 10 1 drops The age of the stalagmite equals: 5.6 x 10 1 drops x (1s/drop) x (1 hr/600 s) x (1 day/ hrs) x (1yr/65 days) = 1.8 x 10 6 yrs

Chapter 17 Solubility and Complex Ion Equilibria

Chem 1046 General Chemistry by Ebbing and Gammon, 8th Edition Prof George W.J. Kenney, Jr Last Update: 27Nov2008 Chapter 17 Solubility and Complex Ion Equilibria These Notes are to SUPPLIMENT the Text,