The dissociation constant of acetic acid at a given temperature is 1.69 ×10^-5.The degree of dissociation of 0.01 M acetic acid in presence of 0.01M HCl is a)0.42 b)0.13 c)0.169×10^-2 d)0.013

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Arun · Answer

Dear student [H 3 0]+ = 0.01 + Ca = 0.01 M [CH 3 COO]- = Ca = 0.01 a [CH 3 COOH] = 0.01 1.69 * 10 -5 = (0.01)(0.01 a)/(0.01) 0.01 a = 1.69 * 10 -5 a = 1.69 * 10 -3 a = 0.169 * 10 -2 Hence option C is correct. Regards Arun (askIITians forum expert)

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The dissociation constant of acetic acid at a given temperature is 1.69 ×10^-5.The degree of dissociation of 0.01 M acetic acid in presence of 0.01M HCl is a)0.42 b)0.13 c)0.169×10^-2 d)0.013

The dissociation constant of acetic acid at a given temperature is 1.69 ×10^-5.The degree of dissociation of 0.01 M acetic acid in presence of 0.01M HCl is a)0.42 b)0.13 c)0.169×10^-2 d)0.013

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The dissociation constant of acetic acid at a given temperature is 1.69 ×10^-5.The degree of dissociation of 0.01 M acetic acid in presence of 0.01M HCl is a)0.42 b)0.13 c)0.169×10^-2 d)0.013

The dissociation constant of acetic acid at a given temperature is 1.69 ×10^-5.The degree of dissociation of 0.01 M acetic acid in presence of 0.01M HCl is a)0.42 b)0.13 c)0.169×10^-2 d)0.013

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