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Solutions to Homework Assignment
CHM 152
Spring 2002
15.22 (a) acidic (b) neutral (c) basic
15.23 The pH can be found using Equation (15.5) of the text.
pH = 14.00 − pOH = 14.00 − 9.40 = 4.
The hydrogen ion concentration can be found as in Example 15.4 of the text. 4.60 = −log[H+]
Taking the antilog of both sides: [H+] = 2.5 × 10 −^5 M
. mL L. mL
mol L
F
HG^
I
KJ
F
HG^
I
KJ^
= 1.98 × 10 − 3 mol KOH
KOH is a strong base and therefore ionizes completely. The OH−^ concentration equals the KOH concentration, because there is a 1:1 mole ratio between KOH and OH−. [OH−] = 0.360 M pOH = −log[OH−] = 0.
15.25 We can calculate the OH−^ concentration from the pOH.
pOH = 14.00 − pH = 14.00 − 10.00 = 4. [OH−] = 10−pOH^ = 1.0 × 10 −^4 M
Since NaOH is a strong base, it ionizes completely. The OH−^ concentration equals the initial concentration of NaOH. [NaOH] = 1.0 × 10 −^4 mol/L
So, we need to prepare 546 mL of 1.0 × 10 −^4 M NaOH.
This is a factor-label problem. We need to perform the following unit conversions. mol/L → mol NaOH → grams NaOH
546 mL = 0.546 L
? g NaOH = 1.0 10 4 mol NaOH 40.00 g NaOH (0.546 L soln) 1 L soln 1 mol NaOH
× −
×^ ×
^
= 2.2 × 10 −^3 g NaOH
15.34 (a) false, they are equal (b) true, find the value of log(1.00) on your calculator (c) true (d) false, if the acid is strong, [HA] = 0.00 M
15.42 Step 1: Calculate the concentration of acetic acid before ionization.
0.0560 g acetic acid × 1 mol acetic acid 60 05. g acetic acid
= 9.33 × 10 −^4 mol acetic acid
9 33. 10 4 mol 0.0500 L soln
× − = 0.0187 M acetic acid
Step 2: Next, recognize that acetic acid is a weak, monoprotic acid. It is not one of the six strong acids, so it must be a weak acid. Express the equilibrium concentrations of all species in terms of initial concentrations and a single unknown x , that represents the change in concentration. Let (− x ) be the depletion in concentration (mol/L) of CH 3 COOH. From the stoichiometry of the reaction, it follows that the increase in concentration for both H+^ and CH 3 COO−^ must be x. Complete a table that lists the initial concentrations, the change in concentrations, and the equilibrium concentrations.
CH 3 COOH ( aq ) U H+^ ( aq ) + CH 3 COO−^ ( aq ) Initial ( M ): 0.0187 0 0 Change ( M ): − x + x + x Equilibrium ( M ): 0.0187 − x x x
Step 3: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant ( K a), solve for x.
K a = [^ ][ [ ]
H CH COO
CH COOH
]
You can look up the K a value for acetic acid in Table 15.3 of your text.
1.8 × 10 −^5 = ( )( ) (.
x x 0 0187 − x )
At this point, we can make an assumption that x is very small compared to 0.0187. Hence, 0.0187 − x ≈ 0.
1.8 × 10 −^5 =
x x 0 0187
Solving for x. x = 5.8 × 10 −^4 M = [H+] = [CH 3 COO − ] [CH 3 COOH] = (0.0187 − 5.8 × 10 −^4 ) M = 0.0181 M
Checking the validity of the assumption, 5 0 0187
.8 × 10 −^4
× 100% = 3.1% < 5%
The assumption is valid.
Step 2: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant ( K a), solve for x.
K a = [^ ][ [ ]
H F
HF
+ − ]
You can look up the K a value for hydrofluoric acid in Table 15.3 of your text.
7.1 × 10 −^4 =
x x 0 60 − x )
At this point, we can make an assumption that x^ is very small compared to 0.60. Hence,
0.60 − x ≈ 0.
Oftentimes, assumptions such as these are valid if K is very small. A very small value of K means that a very small amount of reactants go to products. Hence, x is small. If we did not make this assumption, we would have to solve a quadratic equation.
7.1 × 10 −^4 =
x x 0 60
Solving for x.
x = 0.021 M = [H+]
Step 3: Having solved for the [H+], calculate the percent ionization.
percent ionization = [^ ] × 100% [ ]
H
HF
0 = 0.0^21
M
M
× 100% = 3.5%
(b) − (c) are worked in a similar manner to part (a). However, as the initial concentration of HF becomes smaller, the assumption that x is very small compared to this concentration will no longer be valid. You must solve a quadratic equation.
(b)
(^2 ) a
[H ][F ]
[HF] (0.0046 )
x K x
x^2 + (7.1 × 10 −^4 ) x − (3.3 × 10 −^6 ) = 0 x = 1.5 × 10 −^3 M
Percent ionization =
1.5 ×^ − M
M
× 100% = 33%
(c)
2 4 a
= [H^ ][F ]= = 7.1 10
[HF] (0.00028 )
x x
K
x^2 + (7.1 × 10 −^4 ) x − (2.0 × 10 −^7 ) = 0 x = 2.2 × 10 −^4 M Percent ionization =
× − M
M
× 100% = 79%
As the solution becomes more dilute, the percent ionization increases.
15.51 (a) We construct the usual table.
NH 3 ( aq ) + H 2 O( l ) U NH 4 +( aq ) + OH−( aq ) Initial ( M ): 0.10 0.00 0. Change ( M ): − x + x + x Equilibrium ( M ): (0.10 − x ) x x
2 4 5 b 3
[NH ][OH ] 1.8 10
[NH ] (0.10 )
K x x
×
Assuming (0.10 − x ) ≈ 0.10, we have: (^2 ) 1.8 10
x = × −
x = 1.3 × 10 −^3 M = [OH−]
pOH = −log(1.3 × 10 −^3 ) = 2.
pH = 14.00 − 2.89 = 11.
By following the identical procedure, we can show: (b) pH = 8..
15.53 A pH of 11.22 corresponds to a [H+] of 6.03 × 10 −^12 M and a [OH−] of 1.66 × 10 −^3 M.
The equilibrium is: NH 3 ( aq ) + H 2 O( l ) U NH 4 +( aq ) + OH−( aq )
The concentration of [OH−] = [NH 4 +] (why?) If we let x equal the original concentration of ammonia
5 3 3 b (^3)
−^ −^ − −
× ×
= × =
− ×
K
x
Assuming 1.66 × 10 −^3 is small relative to x , then x = 0.15 M = [NH 3 ]
15.62 For the first stage of ionization:
H 2 CO 3 ( aq ) U H+^ ( aq ) + HCO 3 −^ ( aq ) Initial ( M ): 0.025 0.00 0. Change ( M ): − x + x + x Equilibrium ( M ): (0.025 − x ) x x
1
(^2 ) a 3 2 3
[H ][HCO ]
[H CO ] (0.025 )
x K x
Assuming (0.025 − x ) ≈ 0.025, x = 1.0 × 10 −^4 M
15.78 The salt ammonium chloride completely ionizes upon dissolution, producing 0.42 M [NH 4 +] and 0.42 M [Cl−] ions. NH 4 +^ will undergo hydrolysis because it is a weak acid (NH 4 +^ is the conjugate acid of the weak base, NH 3 ). Step 1: Express the equilibrium concentrations of all species in terms of initial concentrations and a single unknown x , that represents the change in concentration. Let (− x ) be the depletion in concentration (mol/L) of NH 4 +. From the stoichiometry of the reaction, it follows that the increase in concentration for both H 3 O+^ and NH 3 must be x. Complete a table that lists the initial concentrations, the change in concentrations, and the equilibrium concentrations. NH 4 +^ ( aq ) + H 2 O ( l ) U NH 3 ( aq ) + H 3 O+^ ( aq ) Initial ( M ): 0.42 0.00 0. Change ( M ): − x + x + x Equilibrium ( M ): (0.42 − x ) x x
Step 2: You can calculate the K a value for NH 4 +^ from the K b value of NH 3. The relationship is K a × K b = K w or K a =
14 w b^5
= 1.0^10
− −
×
×
K
K
= 5.6 × 10 −^10
Step 3: Write the ionization constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant ( K a), solve for x.
3 3 2 10 a 4
[NH ][H O ]
[NH ] 0.
−
x K x
Assuming (0.42 − x ) ≈ 0.42, x = [H+] = 1.5 × 10 −^5 M pH = −log(1.5 × 10 −^5 ) = 4.
Since NH 4 Cl is the salt of a weak base (aqueous ammonia) and a strong acid (HCl), we expect the solution to be slightly acidic, which is confirmed by the calculation.
