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Question

In which of the following mixed aqueous solutions pH=pKa at equilibrium?


(1)100 ml of 0.1 M CH3COOH+100 ml of 0.1 M CH3 COONa
(2)100 ml of 0.1 M CH3COOH+50 ml of 0.1 M NaOH
(3)100 ml of 0.1 M CH3COOH+100 ml of 0.1 M NaOH
(4)100 ml of 0.1 M CH3COOH+100 ml of 0.1 M NH3

A
(1) is correct
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B
(2) is correct
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C
(3) is correct
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D
Both (1) and (2) are correct
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Solution

The correct option is C Both (1) and (2) are correct
Both (1) and (2) are correct.
pH=pKa at equilibrium for the (1) and (2) aqueous solutions.
(1) A solution containing a weak acid and its salt with strong base is an acidic buffer solution.
pH=pKa+log[CH3COONa][CH3COOH]
pH=pKa+log(0.1×1000.1×100)=pKa (correct).
(2) A solution containing a weak acid and a strong base is an acidic buffer solution.

CH3COOH10 m mol5 m mol+NaOH5 m mol0CH3COO05 m molNa++H2O05 m mol (acidic buffer formed)
pH=pKa+log[CH3COONa][CH3COOH]
pH=pKa+log(55)=pKa (Correct).
(3) A salt of weak acid (acetic acid) and strong base (NaOH) is formed. It undergoes hydrolysis.
CH3COOH10 m mol0+NaOH10 m mol0CH3COO010 m molNa++H2O010 m mol
pH=12(pKw+pKa+logc)=12[14+pKa+log(10200)] (incorrect) (anionic hydrolsis).

(4) Weak acid and weak base give buffer solution.

CH3COOH+NH4OHCH3COO+NH+4
10 mmol 10mmol 0 0
0 0 10mmol 10mmol
In this case, pH won't be equal to pKa because in the weak acid and weak base titration pKb will also come into the picture from the following formula:
pH=12pKw+12pKa12pKb

Hence the correct option is

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