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Q.

0.01 moles of a weak acid HA Ka=2.0×106 is dissolved in 1.0 L of 0.1 M HCl solution. The degree of dissociation of HA is ______×105 (Round off to the Nearest Integer). [Neglect volume change on adding HA. Assume degree of dissociation <<1]

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answer is 2.

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Detailed Solution

ka=H+A[HA]2×106=0.1A0.01A=2×107=2×107=102α  α=2×105=2

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