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.

20mL of 0.2M NaOH are added to 50mL of 0.2M acetic acid (K_(a)=1.85xx1

1.	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.
Answer» Solution :`{:(,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`