Equilibrium
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7.50. The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the PKa of bromoacetic acid.
7.50. The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the PKa of bromoacetic acid.
H+] = cα = 0.1 × 0.132 = 0.0132M
pH = −log [H+] = −log0.0132
= 1.88
The acid dissociation constant is
Ka? = cα2? / (1−α) = 0.1 × (0.132)2 / (1−0.132)
= 2.01×10−3.
pKa? = −logKa? = −log (2.01×10−3) ≈ 2.7
Upvote 
7.49. Calculate the pH of the following solutions:
(a) 2g ofTlOH dissolved in water to give 2 litre of the solution
(b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution
(c) 0.3 g of NaOH dissolved in water to give 200 mL of the solution
(d) l mL of 13.6 M HCl is diluted with water to give 1 litre of the solution.
7.49. Calculate the pH of the following solutions:
(a) 2g ofTlOH dissolved in water to give 2 litre of the solution
(b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of the solution
(c) 0.3 g of NaOH dissolved in water to give 200 mL of the solution
(d) l mL of 13.6 M HCl is diluted with water to give 1 litre of the solution.
(a) For 2g of TlOH dissolved in water to give 2 L of solution:
[TlOH] = [OH−] = (2×1)? / (2×221) = (1 / 221)? M
pOH = −log [OH]− = −log (1/221)?
= 2.35
pH = 14 – pOH = 14 − 2.35 = 11.65
(b) For 0.3 g of Ca (OH)2? dissolved in water to give 500 mL of solution:
[OH−] = 2 [Ca (
Upvote 7.48. Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH.
7.48. Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH.
(a) 0.003 M HCl
[H3? O+] = [HCl] = 0.003M
pH = −log [H+] = −log (3.0×10−3) = 2.523
(b) 0.005 M NaOH
[OH−] = [NaOH] = 0.005M
[H+] = Kw? / [OH−]? = 10−14/ 0.005? =2×10−12
pH= −log [H+]=−log (2×10−12)=11.699
(c) 0.002M HBr
[H+]= [HBr]=0.002
pH= −log [H+]=−log0.002=2.699
(d) 0.002M KOH
[OH−]= [KOH]
Upvote 7.47. It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its PKa.
7.47. It has been found that the pH of a 0.01 M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its PKa.
pH= −log [H+]=4.15
[H+]= antilog (−4.15)= 7.08×10−5
[A−]= [H+]=7.08×10−5
The concentration of undissociated acid is 0.01−0.000071=0.009929M.
HA+H2? O? H3? O++A−
Ka? = [H3? O+] [A−] / [HA]? = (7.08×10−5) (7.08×10−5)? / 0.009929
= 5.05×10−7
pKa? = −logKa? = −log5.05×10−7 ≈ 6.3
Upvote 7.46. The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
7.46. The ionization constant of acetic acid is 1.74 x 10-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
The dissociation equilibrium is
CH3? COOH? CH3? COO− + H+.
Let α be the degree of dissociation.
The equilibrium concentrations of CH3? COOH, CH3COO− and H+ are c (1−α), c (α) and c (α) respectively.
The equilibrium constant expression is Kc? =
Upvote 7.45. The-first ionization constant of H2S is 9.1 x 10-8. Calculate the concentration of HS– ions in its 0.1 M solution and how will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H2S is 1.2 x 10--13, calculate the concentration of S2-under both conditions.
7.45. The-first ionization constant of H2S is 9.1 x 10-8. Calculate the concentration of HS– ions in its 0.1 M solution and how will this concentration be affected if the solution is 0.1 M in HCl also? If the second dissociation constant of H2S is 1.2 x 10--13, calculate the concentration of S2-under both conditions.
(i) To calculate [HS−] in absence of HCl:
Let, [HS−] = x M.
H2? S? H+ + HS−
The initial concentrations of H2? S, H+ and HS− are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-x M, x M and x M respectively.
Ka? = [H2? S] [H+] [HS−]?
9.1×
Upvote 7.44. The ionization constant of phenol is 1.0 x 10-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?
7.44. The ionization constant of phenol is 1.0 x 10-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?
C6H5OH? C6H5O- + H+
| C6H5OH | C6H5O- | H+ |
Initial | 0.05 M | 0 | 0 |
After dissociation | 0.05 –x | x | x |
Ka = x2 / (0.05 - x) = 1.0 x 10-10
=> x2 / 0.05 = 1.0 x 10-10
=> x2 = 5 x 10-12
=> x= 2.2 x 10-6 M
In presence of 0.01 C6H5Na, suppose y is the am
Upvote 7.43. The ionization constant of HF, HCOOH and HCN at 298 K are is 6.8 x 10-4, 1.8x 10-4 and 4.8 x 10-9 respectively, calculate the ionization constant of the corresponding conjugate base.
7.43. The ionization constant of HF, HCOOH and HCN at 298 K are is 6.8 x 10-4, 1.8x 10-4 and 4.8 x 10-9 respectively, calculate the ionization constant of the corresponding conjugate base.
For F–, Kb =Kw/Ka= 10-14/ (6.8 x 10-4)
= 1.47 x 10-11 = 1.5 x 10-11 .
For HCOO-, Kb = 10-14/ (1.8 x 10-4)
= 5.6 x 10-11
For CN–, Kb= 10-14/ (4.8 X 10-9)
= 2.08 x 10-6
Upvote 7.42. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
7.42. The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
pH = – log [H+] or log [H+] = – pH = – 3.76 = 4.24
Therefore, [H+] = Antilog 4.24 = 1.738 x 10-4 = 1.74 x 10-4 M
Upvote 7.41. The concentration of hydrogen ions in a sample of soft drink is 3.8 x 10-3 M. What is the pH value?
7.41. The concentration of hydrogen ions in a sample of soft drink is 3.8 x 10-3 M. What is the pH value?
pH = – log [H+] = – log (3.8 x 10-3) = – log 3.8 + 3 = 3 – 0.5798 = 2.4202 = 2.42
Upvote