Honors Chemistry

Answers to Chapters 15 & 16 Study Questions

1.  a)  strong acid + strong base:  steepest at equivalence, low pH at start depending on concentration of acid, equivalence at pH 7, no buffering.

b) weak acid + strong base:  less steep at equivalence than strong acid, low pH at start depending on Ka and concentration of acid, equivalence at basic pH, buffering.

c) weak base + strong acid:  less steep at equivalence than strong base, high pH at start depending on Kb and concentration of base, equivalence at acid pH, buffering.

2.  a) blue                                 b)  pH =  5

c)  an indicator changes color when [HA] = [A-] and [H+] = Ka;

therefore Ka = [H+] = 10-3.5 = 3.2 × 10-4

d)  weak base + strong acid ® weak acid;   acid endpoint;  methyl red

3.  VA × MA =  VB × MB;  VA × 2.00 M  =  12.5 mL × 0.800 M; VA = 0.500 mL

(optional pH of solution:)  H+ + NH3 ® NH4+; final volume = 12.5 + .5 = 13.0 mL

Final [NH4+]:  12.5 mL × 0.0800 M = 13.0 mL × M2;  M2 = 0.0769 M NH4+ at end.

So, calculate the pH of 0.0769 M NH4+:

Ka = ;  x2 = (5.6 x 10-10)(0.0769) = 4.3 x 10-11;

x = (4.3 × 10-11)½  = 6.6 × 10-6 M;  pH = − log (6.6 × 10-6 M) = 5.2

(you get the same answer if you used 0.080 M NH4+.)

4.  a)  Ka(CH3COOH) = 1.8 × 10-5;  [H+] = 1.0 x 10-5 M

= 1.8

b)  Mix the solutions in a 1.8: 1 ratio.  For example, mix 180. mL of 0.100 M NaCH3COO with 100 mL 0.100 M CH3COOH.

c)  Mix 280 mL 0.100 M CH3COOH and 180. mL 0.100 M NaOH.

d)  Another pH 5.00 buffer might be Al(H2O)63+ or even benzoic acid.

5.  a)  limiting reactant problem:  0.750 L × 0.400 M NaOH = 0.300 moles OH

0.250 L × 0.800 M HCl = 0.200 mol H+.

0.300 moles OH  + 0.200 mol H+ ®  0.200 mol H2O + 0.100 mol OH remain.

Total volume = 250 mL + 750 mL = 1.00 liter.

[OH]  =   = 0.100 M; pOH = 1.0; pH = 13.0

b)

= 6.36 − 0.30  =  6.06

6.  MgCl­2  Mg2+(aq)  +  2  Cl-(aq);    Ksp  =   [Mg2+] × [Cl-]2

[MgCl­2]  =   = 0.778 M

0.778 M MgCl­2:  [Mg2+] = 0.778 M;  [Cl-] = 2(0.778 M) = 1.56 M

Ksp  =   [Mg2+] × [Cl-]2 = (0.778) × (1.56)2 =  1.88

7.  CuCrO­4  Cu2+(aq)  +  CrO42-(aq);    Ksp =  [Cu2+] × [CrO42-]  =  3.6 x 10-6;

x  =  [CuCrO­4]  =  [Cu2+]  =  [CrO42-];  Ksp =  3.6 x 10-6  =  x2

x  =  (3.6 x 10-6)½  =  1.9 x 10-3 M

8.  2 AgNO3(aq)  +  Na2CO3(aq)   ®   2 NaNO3(aq)  +  Ag2CO3(s)

Ag2CO3(s)  2 Ag+(aq)  +  CO32-(aq);  Ksp =  [Ag+]2 × [CO32-]  =  8.1 x 10-12

x = [Ag+];  [CO32-] = 0.020 M;  Ksp =  x2(0.020) = 8.1 x 10-12

= 4.1 x 10-10;  x  =  (4.1 x 10-10)½  =  2.0 x 10-5 M

9.  Pb(NO3)2(aq)  +  2 NaBr(aq)   ®   2 NaNO3(aq)  +  PbBr2(s).

PbBr2(s)   Pb2+(aq) +  2 Br-(aq);    Ksp  =   [Pb2+] × [Br-]2;   Ksp = 4.6 x 10-6

When solutions are mixed in a 1:1 ratio, each is diluted by a factor of 2.

[Pb2+] = (0.0100 M)/2 = 0.00500 M;  [Br-] = (0.0200 M)/2 = 0.0100 M

Q =   [Pb2+] × [Br-]2 = (0.00500) × (0.0100)2  =  5.0 x 10-7

5.0 x 10-7 < 4.6 x 10-6;  Q < Ksp;  therefore, no precipitate forms

10.  a)  strong acid reacts with the weak acid in the buffer:

H+(aq)  + HCO3-(aq)  ®  H2CO3(aq)

b)  Ag3PO4(s)  3 Ag+(aq)  +  PO43-(aq);  Ksp =  [Ag+]3 × [PO42-]

11.  molar mass = mass/moles;

moles acid  =  moles base:  17.5 mL  =  0.00469 moles acid

molar mass  =   = 90.0 g/mol