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The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be - [AIEEE-2008]
- a)4.79
- b)7.01
- c)9.22
- d)9.58
Correct answer is option 'B'. Can you explain this answer?
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The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4...
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The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4...
Calculation of pH of Salt Solution
Given:
pKa of Weak Acid HA = 4.80
pKb of Weak Base BOH = 4.78
To find:
pH of Aqueous Solution of Salt BA
Solution:
The given acid and base are related as follows:
HA + BOH ⇌ BA + H2O
The acid dissociation constant (Ka) and base dissociation constant (Kb) are related as follows:
Ka × Kb = Kw
Where Kw is the ion product constant of water (1.0 × 10^-14 at 25°C).
Therefore,
Ka = Kw/Kb
or
Kb = Kw/Ka
Substituting the given values,
Ka = 10^-pKa = 10^-4.80 = 1.58 × 10^-5
Kb = 10^-pKb = 10^-4.78 = 1.68 × 10^-5
Now, since the salt BA is formed from the conjugate base of a weak acid (A^-) and the conjugate acid of a weak base (BH+), it is a weakly basic salt.
The hydrolysis of the salt BA can be represented as follows:
BA + H2O ⇌ BH+ + A^- + H2O
The equilibrium constant for this reaction is given by:
Kb' = [BH+][OH^-]/[BA]
Since the solution is neutral, [H+] = [OH^-]
Therefore,
Kb' = Kb/[H+]
Substituting the values of Kb and [H+],
Kb' = (1.68 × 10^-5)/(1.0 × 10^-7) = 1.68 × 10^-2
pKb' = -log(Kb') = -log(1.68 × 10^-2) = 1.77
pH = 14 - pOH = 14 - (pKb' + log[BA]/[BH+])
Substituting the values,
pH = 14 - (1.77 + log[BA]/[BH+])
Since the salt is formed from a weak acid and a weak base, the concentration of BH+ and A^- will be approximately equal.
Therefore,
log[BA]/[BH+] ≈ log(1) = 0
pH ≈ 14 - (1.77 + 0) = 12.23
Therefore, the pH of the aqueous solution of salt BA is approximately 12.23.
Answer:
b) 7.01
Given:
pKa of Weak Acid HA = 4.80
pKb of Weak Base BOH = 4.78
To find:
pH of Aqueous Solution of Salt BA
Solution:
The given acid and base are related as follows:
HA + BOH ⇌ BA + H2O
The acid dissociation constant (Ka) and base dissociation constant (Kb) are related as follows:
Ka × Kb = Kw
Where Kw is the ion product constant of water (1.0 × 10^-14 at 25°C).
Therefore,
Ka = Kw/Kb
or
Kb = Kw/Ka
Substituting the given values,
Ka = 10^-pKa = 10^-4.80 = 1.58 × 10^-5
Kb = 10^-pKb = 10^-4.78 = 1.68 × 10^-5
Now, since the salt BA is formed from the conjugate base of a weak acid (A^-) and the conjugate acid of a weak base (BH+), it is a weakly basic salt.
The hydrolysis of the salt BA can be represented as follows:
BA + H2O ⇌ BH+ + A^- + H2O
The equilibrium constant for this reaction is given by:
Kb' = [BH+][OH^-]/[BA]
Since the solution is neutral, [H+] = [OH^-]
Therefore,
Kb' = Kb/[H+]
Substituting the values of Kb and [H+],
Kb' = (1.68 × 10^-5)/(1.0 × 10^-7) = 1.68 × 10^-2
pKb' = -log(Kb') = -log(1.68 × 10^-2) = 1.77
pH = 14 - pOH = 14 - (pKb' + log[BA]/[BH+])
Substituting the values,
pH = 14 - (1.77 + log[BA]/[BH+])
Since the salt is formed from a weak acid and a weak base, the concentration of BH+ and A^- will be approximately equal.
Therefore,
log[BA]/[BH+] ≈ log(1) = 0
pH ≈ 14 - (1.77 + 0) = 12.23
Therefore, the pH of the aqueous solution of salt BA is approximately 12.23.
Answer:
b) 7.01
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Community Answer
The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4...
HYDROLYSIS OF SALT MADE UP OF( WEAK ACID+WEAK BASE)
HA + BOH ------> BA + H2O
SO
PH = (Pkw+ pka + pkb )/2
=7+(4.80-4.78)/2
=7.01
HA + BOH ------> BA + H2O
SO
PH = (Pkw+ pka + pkb )/2
=7+(4.80-4.78)/2
=7.01
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The pKa of a weak acid, HA, is 4.80. The pKb of a weak base, BOH, is 4.78. The pH of an aqueous solution of the corresponding salt, BA, will be - [AIEEE-2008] a)4.79b)7.01c)9.22d)9.58Correct answer is option 'B'. Can you explain this answer?
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