Question

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Answer 1

Answer: 2) 8.87

Explanation:

Moles of CH3COOH = 0.2 x 100 x 1 = 20 mmol

Moles of NaOH = 0.2×100×1=20 mmol

So there is complete neutralisation

[salt]=C= 20/(100+100) = 0.1M

pH= 7+ 1/2 * pKa + 1/2 $log_{10} C$

pH    = 7 + 1/2 * 4.74  + $log(0.1)$ = 8.87

Answer 2

Answer:

Moles of CH3COOH = 0.2 x 100 x 1 = 20 mmol

Moles of NaOH = 0.2×100×1=20 mmol

So there is complete neutralisation

[salt]=C= 20/(100+100) = 0.1M

pH= 7+ 1/2 * pKa + 1/2 log_{10} Clog

10

C

pH = 7 + 1/2 * 4.74 + log(0.1)log(0.1) = 8.87