Equilibrium
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7.60. The pHof 0.1M of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
7.60. The pHof 0.1M of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
c=0.1M
pH= −log [H+]
=> 2.34 = −log [H+]
So, −log [H+]= 2.34
=> [H+]= 4.5×10−3
Also,
[H+]= cα
=>4.5×10−3= 0.1×α
=> α=4.5×10−2= 0.045
Then,
Ka = α2/c = (45×10−3)2/ 0.1
=202.5×10−6
=2.02×10−4
Upvote 
7.59. The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
7.59. The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also?
Let the degree of ionization of propanoic acid be α. Then, representing propionic acid as HA, we have:
HA + H2O ⇔ H3O+ + A−
(.05−0.0α)≈0.5
Upvote 7.58. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
7.58. The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Solubility of Sr (OH)2=19.23g/L
The molecular weight of Sr (OH)2 is 87.6 + 2 (17)=121.6
Then, concentration of Sr (OH)2=19.23 /121.63M=0.1581M
Sr (OH)2 (aq)→Sr2+ (aq)+2 (OH−) (aq)
∴ [Sr2+]=0.1581M
[OH−]=2×0.1581M=0.3126M
Now, Kw= [OH−] [H+]
=> [H+] = 10−14 / 0.3126
=> [H+]=3.2×10−14
Upvote 7.57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
7.57. If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
[KOH]= [K+]= [OH−]= (0.561×1000) / (56×200)? =0.050M
[H+]=Kw / [OH−]? =10−14 / 0.05? =2.0×10−13
pH=−log [H+]=−log (2.0×10−13)
=12.7
Upvote 7.56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
7.56. The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH=−log [H+]
Hence, [H+] = 10−pH
Milk: [H+] = 10−6.8 = 1.58×10−7M
Black coffee: [H+] = 10−5.0 =1×10−5M
Tomato juice: [H+] = 10−4.2 =6.31×10−5M
Lemon juice: [H+]=10−2.2 = 6.
Upvote 7.55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2
(c) Human blood, 7.38 (d) Human saliva, 6.4.
7.55. Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below:
(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2
(c) Human blood, 7.38 (d) Human saliva, 6.4.
a. Human muscle fluid 6.83
pH=6.83
pH=−log [H+]
∴6.83=−log [H+]
[H+]= 1.48 x 10−7M
b. Human stomach fluid, 1.2:
pH=1.2
1.2=−log [H+]
∴ [H+] = 0.063 M = 6.3 x 10-2 M
c. Human blood, 7.38:
pH=7.38=−log [H+]
∴ [H+]= 4.17 x 10−8M
d. Human saliva, 6.4:
pH=6.4
6.4=−log [H+]
[H+]= 3.98 x 10−7 M
Upvote 7.54. The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?
7.54. The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH?
Kb= 5.4×10−4
c= 0.02M
Then, α= (Kb /c)1/2
α= (5.4×10−4 / 2 x 10-2)1/2 =0.1643
(CH3)2NH+H2O ↔ (CH3)2NH+2+OH-
[ (CH3)2NH] = 0.02 – x ≈ 0.02
[ (CH3)2NH+2] = x
[OH-] = 0.1 + x
≈ 0.1
Now, Kb= [ (CH3)2NH+2] [OH−]/ [ (CH3)2NH] = (x × 0.1) / (0.025).
x = 1.08 x 10-4
% of dimethylamine ionised = (1.08 x 10-4) x (10
Upvote 7.53. Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74.
How is the degree of dissociation affected when its solution also contains
(a) 0.01M (b) 0.1M in HCl?
7.53. Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74.
How is the degree of dissociation affected when its solution also contains
(a) 0.01M (b) 0.1M in HCl?
pKa? =? logKa= 4.74
Ka? = 10? pKa =10?4.74 = 1.8*10?5
Let x be the degree of dissociation. The concentration of acetic acid solution, C = 0.05 M
The degree of dissociation,
x= (Ka / C)1/2? = (1.8*10?5 / 0.05)1/2 ? = 0.019
(a) The solution is also 0.01 M in HCl.
Let x M be the hydrogen ion concentr
Upvote 7.52. What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
7.52. What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from Table 7.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Let c be the initial concentration of C6H5NH3+ and x be the degree of ionisation.
C6H5NH2 + H2O? C6H5NH3+ + OH-
c (1-x) cx
Upvote 7.51. The pH of 0.005 M codeine (C18H21N03) solution is 9.95. Calculate the ionization constant and PKb.
7.51. The pH of 0.005 M codeine (C18H21N03) solution is 9.95. Calculate the ionization constant and PKb.
pH = 9.95,
pOH = 14 – pH = 14 − 9.95 = 4.05
[OH−] = 10−pOH = 10−4.05 = 8.913 × 10−5
Codeine + H2? O? CodeineH+ + OH−
The ionization constant, Kb? = [CodeineH+] [OH−] / [codeine]?
= [ (8.913×10−5)× (8.913×10−5)] / 5×10−3
= 1.588×10−6.
pKb? = −log (1.588×10−6)
= 5.8
Upvote