Chemistry, Third edition

ANSWERS TO EXERCISES AND REVISION QUESTIONS

(iii)C 6 H 5 NH 2 (aq)H 2 O(l) C 6 H 5 NH 3 +(aq)OH–(aq)

conjugate acid

[H 3 O+(aq)] 10 – 7.801.6 10 –^8 mol dm–^3

[OH–(aq)] 1.0^10 –^14

1.6 10 –^8 6.3^10

– (^7) mol dm– 3
CBinitial conc [OH

• (aq)] 2
  Kb(T)

[6.3^10

– (^7) ] 2
4.3 10 –^10 9.2^10 –^4 mol dm–^3
16I
Salt Formula Parent acid Acidic Basic Neutral
and base
Iron(III) Fe(NO 3 ) 3 HNO 3 (SA) 
nitrate Fe(OH) 3 (WB)
Calcium CaCl 2 HCl (SA) 
chloride Ca(OH) 2 (SB)
Sodium Na 2 SO 4 H 2 SO 4 (SA) 
sulfate NaOH(SB)
Ammonium C 6 H 5 CO 2 NH 4 C 6 H 5 CO 2 H (WA)
benzoate NH 3 (aq) (WB)
(Ka (benzoic acid) Kb (ammonia)).
16J
Measure pH at different acid and salt concentrations.
Plot [H 3 O+(aq)] against the ratio CA:CS. The slope of the
graph equals Ka(T).
16K
(i)Number of moles of ethanoic acid 0.10 25/1000
2.5 10 –^3.
Number of mol of sodium ethanoate 0.10 50/1000
5.0 10 –^3.
Concentration of ethanoic acid (2.5 10 –^3 )/75/
1000 3.3 10 –^2 mol dm–^3 CA.
Concentration of sodium ethanoate (5.0 10 –^3 )/
75/10006.6 10 –^2 mol dm–^3 CS.
CA/CS0.50
[H 3 O+(aq)] CACKa(T)
s
0.501.8 10 –^5 9.0 10 –^6 mol dm–^3
pHlog (9.0 10 –^6 )5.05
(ii)Similar calculations show that if CA:Cs2.0,
[H 3 O+(aq)] CACKa(T)
s
2.0 1.8 10 –^5 3.6 10 –^5 mol dm–^3
pHlog (3.6 10 –^5 )4.44
16L
Cs/CA 1
pHpK Cs
alogCA4.180, so pKa4.18
Ka 10 – 4.186.6 10 –^5 mol dm–^3. This suggests that
the acid is benzoic acid.
16M
pH Phenolphthalein Bromocresol green
12 colourless yellow
13.7 colourless yellow
15 colourless blue/yellow
16 colourless blue
18 colourless blue
11 pink blue
16N
(i)Alizarin yellow and (to a lesser extent) methyl orange.
(ii)Pairs (a) and (b); (c) involves a weak base.
16O
(i)Bromocresol green changes over the pH range 4.0–
5.6. This is too soon, and its use in the titration of a
weak acid and a strong base will lead to an
underestimation of the equivalence point.
(ii)For the mixture in question, 12.50 cm^3 represents
half-equivalence, i.e. the volume of NaOH required to
neutralize half the ethanoic acid. At half-equivalence the
concentration of ethanoic acid equals the concentration
of sodium ethanoate, i.e. CsCAand log CS/CA0.
The Henderson–Hasselbalch equation then simplifies to
pH of solution pKa. This gives us an approximate
method (in addition to the method outlined in Exercise
16J) of finding the acidity constant of a weak acid. For
example, Fig. 16.3(b) shows that after the addition of
12.5 cm^3 of NaOH, pH 4.7 pKa(CH 3 CO 2 H).
Therefore,
Ka
2  10 –^5 mol dm–^3.
16P
(i)From Table 16.2, Ka4.3 10 –^7 mol dm–^3. Applying
the equation
Ka(T) [H^3 O
+(aq)] 2
[CO 2 (aq)]
(you may recognize this as equation (16.5).
[H 3 O+(aq)]^2 4.3 10 –^7 1.2 10 –^5
5.2 10 –^12 mol^2 dm–^6
[H 3 O+(aq)] (5.2 10 –^12 ) 2.3 10 –^6
pH log(2.3 10 –^6 )5.62
(ii)Salt occupies some of the space that would
otherwise be taken up by dissolved carbon dioxide.
Therefore, salty water is less acidic than deionized
water.
Revision questions
16.1 (i)F
(ii)F (A pH of 6 would cause it to be yellow),
(iii)T, (iv)T, (v)T.
16.2 (i)HNO 3 provides 0.080 mol dm–^3 of H+(aq).
pHlog (0.080) 1.10
(ii)KOH provides 0.080 mol dm–^3 of OH–(aq) in solution.
[H+(aq)]1.0^10

• 14
  0.080 1.3^10 –^13 mol dm–^3
  pHlog[1.3 10 –^13 ]12.89
  16.3 (i)Table 16.1 shows that Kw51.3 10 –^14 mol^2
  dm–^6 at 100 °C. In pure water [OH–(aq)][H 3 O+(aq)];
  therefore 51.3 10 –^14 [H 3 O+(aq)]^2 or [H 3 O+(aq)]
  7.16 10 –^7 mol dm–^3.
  pH (100 °C) log[7.16 10 –^7 ]6.145
  (ii)Number of moles of NaOH [11.0/1000) 0.040
  4.4 10 –^4.
  Number of moles of HCl (9.0/1000) 0.0403.6
  10 –^4.
  Since 1 HCl 1 NaOH, after neutralization we have 0.80
   10 –^4 mol of OH–(aq) spread out over 20 cm^3 of
  solution. This is a concentration of
  (0.80 10 –^4 )/20/10004.0 10 –^3 mol dm–^3 of
  [OH–(aq)]
  At 25 °C,
  [H+(aq)]1.0^10


• 14
  4.0 10 –^3 2.5^10 –^12 mol dm–^3
  pHlog[2.5 10 –^12 ]11.60
  16.4 (i)
  C 17 H 19 O 3 N(aq)H 2 O(l) C 17 H 19 O 3 NH+(aq)
  OH–(aq) (A)
  Since:
  Kb1.6 10 –^6 mol dm–^3 ;
  M(C 17 H 19 O 3 N)285 g mol–^1 ;
  1.0 mg 0.0010 g

Number of mols of morphine 0.0010 g / 285 g mol–^1

3.5 10 –^6 mol.

Concentration of morphine CB3.5 10 –^6 mol per

dm^3

We now use the equation

Kb(T) [OH

• (aq)] 2
  CB
  Rearranging and substituting values,
  [OH–(aq)] (Kb(T)CB)
   (3.5 10 –^6 1.6 10 –^6 )
  2.4 10 –^6 mol dm–^3
  pOHlog (2.4 10 –^6 )5.62
  Also, since pH pOH14,
  pH 14 5.628.38
  (ii)The OH–(aq) produced in reaction (A) reacts with the
  H 3 O+(aq) from HCl(aq):
  H 3 O+(aq)OH–(aq)2H 2 O(l) (B)
  Addition of equations (A) and (B) gives the overall
  reaction
  C 17 H 19 O 3 N(aq)H 3 O+(aq) C 17 H 19 O 3 NH+(aq)
  H 2 O(l)
  The number of mols of morphine in 100 cm^3 of
  solution is 3.5 10 –^7. This requires an equal number
  of moles of HCl(aq) for neutralization. We calculate
  the volume of 0.0010 mol dm–^3 HCl(aq) required as
  follows:

volume of HCl(aq)  moles 3.5 ^10 –^7

concentration 0.0010

3.5 10 –^4 dm^3 or 0.35 cm^3

16.5The ionization of phenol is represented as

C 6 H 5 OH(aq)H 2 O(l) C 6 H 5 O–(aq)H 3 O+(aq)

using [H 3 O+(aq)] (KaCA),CA5.00 10 –^3 mol

dm–^3 andKa1.3 10 –^10 mol dm–^3 (from Table 16.2),

[H 3 O+(aq)] (1.3 10 –^10 5.00 10 –^3 )

8.1 10 –^7 mol dm–^3

pHlog[H 3 O+(aq)]log(8.06 10 –^7 )6.094

16.6The acidity constant for NH 4 +(aq) is the equilibrium

constant that applies to the equation

NH 4 +(aq)H 2 O(l) NH 3 (aq)H 3 O+(aq)

that is,

Ka(NH 4 +(aq))[NH^3 (aq)][H^3 O

+(aq)]

[NH 4 +(aq)]

For the equation

NH 3 (aq)H 2 O(l) NH 4 +(aq)OH–(aq)

the following is true:

Kb(NH 3 (aq))[NH^4

+(aq)][OH–(aq)]

[NH 3 (aq)]

We also know that

Kw[H 3 O+(aq)][OH–(aq)]

The relationship

KaKbKw (A)

is now proved by substitution and cancellation of

terms:

[NH 3 (aq)][H 3 O+(aq)][NH+ 4 (aq)][OH–(aq)] [H

[NH 4 +(aq)] [NH 3 (aq)]^3 O+(aq)][OH–(aq)]

Ka  Kb  Kw

Equation (A) applies for all pairs of conjugate acids and

bases. Here,

Ka(NH 4 +(aq) ) Kw/Kb(NH 3 (aq))

1.0 10 –^14 /(1.8 10 –^5 )

5.6 10 –^10 mol dm–^3

16.7 (i)Assuming that the dissolved CO 2 concentration

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