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If 8.2g of anhydrous sodium acetate is added to 11 of 0.l(M) acetic acid solution, then what will be the change in degree of ionisation of the acid? [Given: K_1(CH_3COOH)=1.8xx10^(-5)] |
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Answer» SOLUTION :If the degree of dissociation of monobasic acid, HA in its aqueous solution `=prop`, MOLAR concentration=c(M) and dissociation constant of the acid`=K_a`, then `K_a=(prop^2c)/(1-prop)` Let the solution is x times diluted. Hence, `c=(0.1)/(x)(M)` As given, degree of ionisation of HA in the dilute solution`=2prop` `:.2prop=sqrt((K_a)/((0.1//x)))=sqrt((x xxK_a)/(0.1))` From [1] & [2], `2xxsqrt((K-a)/(0.1))=sqrt((x xxK-a)/(0.1))` or `sqrtx=2` `:.x=4` Therefore, if the solution is diluted 4 times, then degree of dissociation of HA in the diluted solution will be 2 times that of the initial solution. |
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