In 20,, ml ,,0.4 ,M-HA solution, 80,, ml water is added. Assuming volume to be additive, pH of

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6-2.Equilibrium-II (Ionic Equilibrium)

hard

In $20\,\, ml \,\,0.4 \,M-HA$ solution, $80\,\, ml$ water is added. Assuming volume to be additive, the $pH$ of final solution is

$(K_a \,\,of\,\, HA = 4 \times 10^{-7} ,\, log\,2 = 0.3)$

A

$4.30$

B

$3.75$

C

$3.40$

D

$3.70$

Solution

${[{\text{HA}}]_{final}}_{\text{ }} = \frac{{20 \times 0.4}}{{100}} = 0.08\,{\text{M}}$

$\therefore \left[ {{{\text{H}}^ + }} \right] = \sqrt {{{\text{K}}_{\text{a}}} \cdot {\text{C}}} $

$ = \sqrt {4 \times {{10}^{ – 7}} \times 0.08} = \sqrt {32 \times {{10}^{ – 9}}} {\text{ }}$

${\text{and pH}} = – \log {\left( {32 \times {{10}^{ – 9}}} \right)^{1/2}} = 3.75$

Standard 11

Chemistry

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