InterviewSolution
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InterviewSolution
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`20mL` of `0.2M NaOH` are added to `50mL` of `0.2M` acetic acid `(K_(a)=1.85xx10^(-5))` Take `log2=0.3,log3=0.48` (1) What is `pH` of solution? (2) Calculate volume of `0.2M NaOH` required to make the `pH` of origin acetic acid solution.`4.74`. |
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Answer» `{:(,NaOH+,CH_(3)COOHrarr,CH_(3)COONa+,H_(2)O),("Millimole added",20xx0.2,50xx0.2,,),(,=4,=10,0,0),("Millimole after reaction",0,6,4,4):}` `{"Molarity"}=("millimole")/("Total volume")` `[CH_(2)COOH]=(6)/(70)` &`[CH_(2)COONa]=(4)/(70)` rArr Buffer solution consisting of a weak acid & its salt with a strong base. (2) Let `V mL` of `0.2M NaOH` is required to make `pH=4.74`,Then `NaOH` should be completely used up (:. final solution is required to be acidic) `{:(,NaOH+,CH_(3)COOHrarr,CH_(3)COONa+,H_(2)O),("Millimole added",0.2xxV,50xx0.2,,),(,=0.2V,=10,0,0),("Millmole after reaction",0,(10-2V),0.2V,0.2V):}` `:.["Acid"]=(10-1.2V)/(50+V),["Salt"]=(0.2V)/(50+V)` `:.4.74=-log1.85xx10^(-5)+log""((0.2V)(50+V))/((10-0.2V)(50+V))` `:.V=25mL` |
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