`50 ml` of `0.05 M Na_(2)CO_(3)` is titrated against `0.1 M HCl`. On adding `40 ml` of `HCl, pH ` of the resulting solution will be `[{:(H_(2)CO_(3):pK_(a_(1))=6.35,pK_(a_(2))=10.33),(log3=0.477,log2=0.30):}]`A. `6.35`B. `6.526`C. `8.34`D. `6.173`

`50 ml` of `0.05 M Na_(2)CO_(3)` is titrated against `0.1 M HCl`. On a

1.	`50 ml` of `0.05 M Na_(2)CO_(3)` is titrated against `0.1 M HCl`. On adding `40 ml` of `HCl, pH ` of the resulting solution will be `[{:(H_(2)CO_(3):pK_(a_(1))=6.35,pK_(a_(2))=10.33),(log3=0.477,log2=0.30):}]`A. `6.35`B. `6.526`C. `8.34`D. `6.173`
Answer» Correct Answer - D `CO_(3)^(2-) + H^(-) rArr HCO_(3)^(-)` Initial millies moles `50 xx 0.05 40 xx 0.1` Final milli moles `1.5 , 2.5` `{:(,HCO_(3)^(-)+,H^(+),rarr,H_(2)CO_(3)),("Initial milli mles",2.5,1.5,,"__"),("Final milli moles",1,"___",,1.5):}` `pH = pK_(a_(1)) + "log"([HCO_(3)^(-)])/([H_(2)CO_(3)]) = 6.173`

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`50 ml` of `0.05 M Na_(2)CO_(3)` is titrated against `0.1 M HCl`. On adding `40 ml` of `HCl, pH ` of the resulting solution will be `[{:(H_(2)CO_(3):pK_(a_(1))=6.35,pK_(a_(2))=10.33),(log3=0.477,log2=0.30):}]`A. `6.35`B. `6.526`C. `8.34`D. `6.173`

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