A one litre solution contains 0.08 mole of acetic acid (K_a = 1.75 × 10^–5). To this solution, 0.02 mole of NaOH is added. Then the pH of resulting solution is [log 1.75 = 0.243]

Easy

Solution

Following reaction will occur

CH₃COOH
$\large +$
NaOH

$\large \rightarrow$ CH₃COONa $\large +$ H₂O

t = 0 0.08 0.02 0 0

$\large t=t_\infty$ 0.08 - 0.02 0.02 - 0.02 = 0 0.02 0

=0.06 = 0

P^H of an acidic buffer is given as
$\large {P^H} = {P^{{K_a}}} + \log \frac{{\left[ {salt} \right]}}{{\left[ {acid} \right]}}$
$\large {P^H} = 4.74 + \log \frac{{0.02}}{{0.06}} = 4.28$

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SCAN ME

	CH₃COOH	$\large +$	NaOH	$\large \rightarrow$ CH₃COONa $\large +$ H₂O
t = 0	0.08		0.02	0 0
$\large t=t_\infty$	0.08 - 0.02		0.02 - 0.02 = 0	0.02 0
	=0.06		= 0