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The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is- [AIEEE-2007]
- a)4.5
- b)2.5
- c)9.5
- d)7.0
Correct answer is option 'C'. Can you explain this answer?
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The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered sol...
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The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered sol...
Explanation:
Step 1: Determine the equilibrium expression
The ionization of the weak acid HA can be represented by the following equilibrium reaction:
HA ⇌ H+ + A-
The equilibrium expression for this reaction can be written as:
Ka = [H+][A-] / [HA]
Step 2: Use the given pKa value to find Ka
The pKa value is given as 4.5. The pKa is defined as the negative logarithm of the acid dissociation constant (Ka). Therefore, we can find Ka by taking the antilog of the pKa value:
Ka = 10^(-pKa) = 10^(-4.5)
Step 3: Calculate the concentration of the weak acid and its conjugate base
Since 50% of the acid is ionized, we can assume that the concentration of the weak acid (HA) is equal to the concentration of its conjugate base (A-). Let's assume the initial concentration of HA is x. Therefore, the initial concentration of A- is also x.
Step 4: Set up an ICE table
An ICE table is used to organize the initial, change, and equilibrium concentrations of the species involved in a chemical reaction:
| | HA | H+ | A- |
|Initial | x | 0 | 0 |
|Change | -x | +x | +x |
|Equilibrium | 0 | x | x |
Step 5: Substitute the equilibrium concentrations into the equilibrium expression
Since the concentration of HA is 0 at equilibrium, the equilibrium expression becomes:
Ka = [H+][A-] / 0
This implies that [H+][A-] = 0
Step 6: Calculate the pOH
The pOH can be calculated using the equation:
pOH = -log10 [OH-]
Since [OH-] is equal to the concentration of A-, which is x, we can substitute x into the equation and solve for pOH:
pOH = -log10 x
Step 7: Find x using the quadratic formula
Since [H+][A-] = 0, we know that either [H+] or [A-] must be zero. However, we cannot assume that [A-] is zero because it is formed as a result of the ionization of HA. Therefore, [H+] must be zero.
Using the equation for the ionization of water, we can write:
Kw = [H+][OH-] = 1.0 x 10^-14
Since [H+] = 0, we can solve for
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The pKa of a weak acid (HA) is 4.5. The pOH of an aqueous buffered solution of HA in which 50% of the acid is ionized is- [AIEEE-2007] a)4.5b)2.5c)9.5d)7.0Correct answer is option 'C'. Can you explain this answer?
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