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A 0.01 M solution of acetic acid is 1.34 % ionised (degree of dissociation = 0.0134) at 298 K. What is the ionization constant of acetic acid ? |
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Answer» Correct Answer - `1.80xx10^(-6)` `{:(,CH_(3)CO OH,hArr,CH_(3)CO O^(-) , +,H^(+),,),("Initial conc.",C "mol" L^(-1),,0,,0,,),("At. eqm.",c(1-alpha),, C alpha,,C alpha,,):}` `K=(C alpha.C alpha)/(C (1-alpha))=(C alpha^(2))/(1-alpha)=(0.01xx(0.0134)^(2))/(1-00.01)=1.8xx10^(-6)` |
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