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Question

500 ml of 0.2 M HCl,500 ml of 0.2 M CH3COOH and 6 g of NaOH is present in a container.The pH of solution at 250C, if pKa of CH3COOH is 4.74

A
4.47
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B
4.74
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C
5.20
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D
6.2
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Solution

The correct option is B 4.74
Writing separate equations for the neutralisation of acids.
HCl+NaOHNaCl+H2O
CH3COOH+NaOHCH3COONa+H2O
Since, HCl is a strong acid, therefore, it will neutralise readily.
Moles of HCl=Molarity×Volume=0.2×0.5=0.1mole
To neutralise 0.1 mole of HCl, 0.1 mole of NaOH is required.
Mass of NaOH required =moles×molar mass of NaOH=0.1×40=4g
Mass of NaOH left after complete neutralisation of HCl=64=2g
Moles of NaOH left =240=0.05moles
So, 0.05 moles of NaOH react with 0.05 moles of CH3COOH to form 0.05 moles of salt .
Moles of CH3COOH initially =0.2×0.5=0.1mole
Therefore, moles of CH3COOH left at equilibrium =0.10.05=0.05
pH=pKa+log[salt][acid]=4.74+log[0.05][0.05]
pH=4.74

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