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pH the Power of H

50 mL of 2N...

Question

$50 m L$ of $2 N$ acetic acid mixed with $10 m L$ of $1 N$ sodium acetate solution will have an approximate $p H$ of: $(K_{a} = 10^{- 5})$ .

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Solution

The correct option is A $4$
According to given data,
Eq.moles of acetic acid $= 0.05 \times 2 = 0.1$
Eq.moles of sodium acetate $= 0.01 \times 1 = 0.01$

On substituting above values in Henderson-Hasselbalch equation
$p H = p K_{a} + l o g \frac{[s a l t]}{[a c i d]}$
$p H = - l o g (10^{- 5}) + l o g \frac{0.01}{0.1}$

Thus, the pH of resulting solution = 4

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