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A weak acid HA after teratment with 12 mL of `0.1M` strong base BOH has a pH of 5. At the end point, the volume of same base required is `26.6mL.K_(a)` of acid is:A. `1.8xx10^(-5)`B. `8.2xx10^(-6)`C. `1.8xx10^(-6)`D. `8.2xx10^(-5)` |
Answer» Correct Answer - B For neutralisation, Total Meq.of acid = Meq.of base `= 26.6xx0.1=2.66` Now for partial netralisation of acid, `{:(,HA+,BOHrarr,BA+,H_(2)O),("Meq. before reaction",2.66,1.2,0,0),("Meq. after reaction",1.46,0,1.2,1.2):}` The rersultant mixture acts as a buffer and, [HA] ans [BA] may be placed in terms of Meq. since volume of mixture is constant. `pH= -log K_(a)+log([Sal t])/([Acid])` or `5= -log K_(a)+log (1.2)/(1.46)` `K_(a)=8.219xx10^(-6)` |
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