5.
(E) Answer (b), NH 3 , is correct. HC2 H 3O 2 will react most completely with the strongest
base. NO3− and Cl − are very weak bases. H 2 O is a weak base, but it is amphiprotic,
acting as an acid (donating protons), as in the presence of NH 3 . Thus, NH 3 must be the
strongest base and the most effective in deprotonating HC2 H 3O 2 .
10.
(M) Again, all of the solutes are strong acids or strong bases.
+
(a) ⎡ H O+ ⎤ = 0.0045 M HCl × 1 mol H3O = 0.0045 M
⎣ 3 ⎦
1 mol HCl
pH = −log ( 0.0045 ) = 2.35
(b)
1 mol H 3O +
⎡⎣ H 3O + ⎤⎦ = 6.14 ×10−4 M HNO3 ×
= 6.14 ×10−4 M
1 mol HNO
3
pH = −log ( 6.14 ×10
(c)
−4
) = 3.21
1 mol OH −
⎡⎣OH − ⎤⎦ = 0.00683 M NaOH ×
= 0.00683M
1 mol NaOH
pOH = −log ( 0.00683) = 2.166 and pH = 14.000 − pOH = 14.000 − 2.166 = 11.83
(d)
2 mol OH −
⎡⎣OH − ⎤⎦ = 4.8 ×10−3 M Ba ( OH )2 ×
= 9.6 ×10−3 M
1 mol Ba ( OH )
2
pOH = −log ( 9.6 ×10
19.
−3
) = 2.02
pH = 14.00 − 2.02 = 11.98
(M) Here we determine the amounts of H 3O + and OH − and then the amount of the one
that is in excess. We express molar concentration in millimoles/milliliter, equivalent to
mol/L.
50.00 mL ×
0.0155 mmol HI 1 mmol H 3O +
×
= 0.775 mmol H 3O +
1 mL soln
1 mmol HI
0.0106 mmol KOH 1 mmol OH −
75.00 mL ×
×
= 0.795 mmol OH −
1 mL soln
1 mmol KOH
The net reaction is H3O+ ( aq ) + OH− ( aq ) → 2H2O(l) .
There is an excess of OH − of ( 0.795 − 0.775 = ) 0.020 mmol OH − .
Thus, this is a basic solution. The total solution volume is ( 50.00 + 75.00 = ) 125.00 mL.
720
Chapter 16: Acids and Bases
0.020 mmol OH −
= 1.6 × 10−4 M, pOH = −log (1.6 × 10−4 ) = 3.80,
125.00 mL
⎡⎣OH − ⎤⎦ =
25.
pH = 10.20
(M) We base our solution on the balanced chemical equation.
H3O+ = 10−1.56 = 2.8 × 10−2 M
−
+
CH 2 FCOOH ( aq ) + H 2 O(l) áà àÜ
àà CH 2 FCOO ( aq ) + H3 O ( aq )
Equation :
Initial :
Changes :
Equil :
Ka =
33.
0.318 M
− 0.028 M
−
−
0M
+ 0.028 M
0.290 M
−
0.028 M
H 3O + CH 2 FCOO −
CH 2 FCOOH
=
b0.028gb0.028g= 2.7 × 10
≈0M
+ 0.028 M
≈ 0.028 M
−3
0.290
(M) Here we determine H 3O+ which, because of the stoichiometry of the reaction,
equals ⎡ C 2 H 3O 2 − ⎤ . ⎡⎣ H 3O + ⎤⎦ = 10− pH = 10−4.52 = 3.0 ×10−5 M = ⎡C 2 H 3O 2 − ⎤
⎣
⎦
⎣
⎦
We solve for S , the concentration of HC2 H 3O 2 in the 0.750 L solution before it dissociates.
Equation:
Initial:
Changes:
Equil:
HC2 H3 O2 (aq) +
SM
−3.0 × 10−5 M
( S − 3.0 ×10−5 ) M
H 2 O(l)
—
—
—




−
C2 H3 O2 (aq)
0M
+3.0 × 10−5 M
3.0 × 10−5 M
+
H 3 O + (aq)
≈ 0M
+3.0 × 10−5 M
3.0 × 10−5 M
[H O + ][C 2 H 3O 2 ]
(3.0 ×10−5 ) 2
Ka = 3
= 1.8 ×10−5 =
[HC2 H 3O 2 ]
( S − 3.0 ×10−5 )
−
−5 2
(3.0 ×10 )
= 1.8 ×10 −5 ( S − 3.0 ×10 −5 ) = 9.0 ×10 −10 = 1.8 ×10 −5 S − 5.4 ×10 −10
9.0 × 10−10 + 5.4 × 10−10
S=
= 8.0 × 10−5 M Now we determine the mass of vinegar needed.
−5
1.8 × 10
8.0 ×10−5 mol HC2 H3O2 60.05 g HC2 H3O2 100.0 g vinegar
mass vinegar = 0.750 L ×
×
×
1 L soln
1 mol HC2 H3O2 5.7 g HC2 H3O2
= 0.063 g vinegar
39.
(M) Let us first compute the H 3O+ in this solution.
721
Chapter 16: Acids and Bases
Equation: HC3 H5 O2 (aq) +
Initial:
0.45 M
Changes: − x M
Equil:
( 0.45 − x ) M




H 2 O(l)
—
—
—
H3O+ (aq)
≈0M
+x M
xM
−
+ C3 H5 O2 (aq)
≈0M
+x M
xM
⎡⎣ H 3 O + ⎤⎦ ⎡C3 H 5 O 2 − ⎤
x2
x2
⎣
⎦
−4.89
−5
Ka =
=
= 10
= 1.3 × 10 ≈
⎡
HC3 H 5 O 2 ⎤⎦⎥
0.45 − x
0.45
⎣⎢
x = 2.4 × 10−3 M ; We have assumed that x = 0.45 M , an assumption that clearly is correct.
46.
(a)
⎡⎣ H3O+ ⎤⎦
2.4 ×10−3 M
equil
α = ⎡
=
= 0.0053 = degree of ionization
HC3H5O2 ⎤⎦⎥ initial
0.45 M
⎣⎢
(b)
% ionization = α × 100% = 0.0053 × 100% = 0.53%
(E) The main estimate involves assuming that the mass percents can be expressed as 0.057
g of 75% H 3 PO4 per 100. mL of solution and 0.084 g of 75% H 3 PO4 per 100. mL of
solution. That is, that the density of the aqueous solution is essentially 1.00 g/mL. Based
on this assumption, the initial minimum and maximum concentrations of H 3 PO4 is:
75g H3PO4
1mol H3PO4
×
100g impure H3PO4 98.00 g H3PO4
= 0.0044 M
1L
100 mLsoln ×
1000 mL
0.057 g impure H3PO4 ×
[H3PO4 ] =
75g H3PO4
1mol H3PO4
×
100g impure H3PO4 98.00 g H 3PO 4
= 0.0064 M
1L
100 mLsoln ×
1000 mL
0.084g impure H 3PO 4 ×
[H3PO4 ] =
Equation:
H 3PO 4 (aq)
+
Initial:
Changes:
Equil:
0.0044 M
−x M
( 0.0044 − x ) M
H 2O(l)




—
H 2 PO−4 (aq)
0M
+x M
x M
—
—
⎡ H 2 PO 4 − ⎤ ⎡ H 3O + ⎤
x2
⎦
⎦ ⎣
K a1 = ⎣
=
= 7.1×10−3
⎡
⎤
H
PO
0.0044
−
x
4 ⎦⎥
⎣⎢ 3
+
H 3O + (aq)
≈0M
+x M
x M
x 2 + 0.0071x − 3.1×10 −5 = 0
−b ± b2 − 4ac −0.0071 ± 5.0 ×10−5 +1.2 ×10−4
x=
=
= 3.0 ×10−3 M = ⎡⎣H3O+ ⎤⎦
2a
2
722
Chapter 16: Acids and Bases
The set-up for the second concentration is the same as for the first, with the exception that
0.0044 M is replaced by 0.0064 M.
⎡ H 2 PO 4 − ⎤ ⎡ H3O + ⎤
x2
⎦
⎦ ⎣
K a1 = ⎣
=
= 7.1×10−3
x 2 + 0.0071x − 4.5 ×10−5 = 0
⎡
⎤
H
PO
0.0064
−
x
4 ⎦⎥
⎣⎢ 3
−b ± b2 − 4ac −0.0071 ± 5.0 ×10−5 +1.8 ×10−4
x=
=
= 4.0 ×10−3 M = ⎡⎣H3O+ ⎤⎦
2a
2
The two values of pH now are determined, representing the pH range in a cola drink.
pH = −log (3.0 ×10−3 ) = 2.52
53.
pH = −log ( 4. ×10−3 ) = 2.40
(E) Protonated codeine hydrolyzes water according to the following reaction:
C18 H 21O3 NH + + H 2O É C18 H 21O3 N + H3O +
pK a = 6.05
pK b = 14 − pK a = 7.95
67.
(E)
(a)
HI is the stronger acid because the H — I bond length is longer than the H — Br
bond length and, as a result, H — I is easier to cleave.
(b)
HOClO is a stronger acid than HOBr because
(i) there is a terminal O in HOClO but not in HOBr
(ii) Cl is more electronegative than Br.
(c)
H 3CCH 2 CCl 2 COOH is a stronger acid than I 3CCH 2 CH 2 COOH both because Cl is
more electronegative than is I and because the Cl atoms are closer to the acidic
hydrogen in the COOH group and thus can exert a stronger e withdrawing effect on
the O−H bond than can the more distant I atoms.
−
72. (E)
(a) SOI2 is the acid, and BaSO3 the base.
(b) HgCl3 – is the acid, Cl– the base.
723