In 20,, ml ,,0.4 ,M-HA solution, 80,, ml wate | vedclass

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Question

${[{	ext{HA}}]_{final}}_{	ext{ }} = \frac{{20 	imes 0.4}}{{100}} = 0.08\,{	ext{M}}$

$	herefore \left[ {{{	ext{H}}^ + }} ight] = \sqrt {{{

Vedclass · Accepted Answer

$3.75$

Vedclass · Answer

${[{	ext{HA}}]_{final}}_{	ext{ }} = \frac{{20 	imes 0.4}}{{100}} = 0.08\,{	ext{M}}$

$	herefore \left[ {{{	ext{H}}^ + }} ight] = \sqrt {{{	ext{K}}_{	ext{a}}} \cdot {	ext{C}}} $

$ = \sqrt {4 	imes {{10}^{ - 7}} 	imes 0.08}  = \sqrt {32 	imes {{10}^{ - 9}}} {	ext{ }}$

${	ext{and pH}} =  - \log {\left( {32 	imes {{10}^{ - 9}}} ight)^{1/2}} = 3.75$

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

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

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