50 ml of 0.05 M Na_(2)CO_(3) is titrated against 0.1 M HCl. On adding 40 ml of HCl, pHof 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):}]

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

1.	50 ml of 0.05 M Na_(2)CO_(3) is titrated against 0.1 M HCl. On adding 40 ml of HCl, pHof 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):}]
Answer» `6.35` `6.526` `8.34` `6.173` Solution :`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`