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The conductivity of 0.001M acetic acid is 4 times 10^-8 S//cm. Calculate the dissociation constant of acetic acid, if molar conductivity at infinite dilution for acetic acid is 390 S cm^2//mol. |
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Answer» SOLUTION :`A_m=(K TIMES 1000)/m` `=(4 times 10^-5 times 1000)/0.001` `alpha=A_m/A_m^0=40/390=0.102` |
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