1.

50mL of 0.05M Na_(2)CO_(3) is titrated against 0.1 M HCl ,pH of the solution will be [Given :For H_(2)CO_(3) ,pK_(a)=6.35,pK_(a)=10.33]

Answer»

`6.35`
`6.526`
`8.34`
`6.173`

Solution :`{:(,CO_(3)^(2-),+,H^(+),rarr,HCO_(3)^(-),),("INITIAL milli-moles",50xx0.05,,40xx0.1,,-,),("FINAL milli-moles",-,,1.5,,2.5,),(,HCO_(3)^(-),+,H^(+),rarr,H_(2)CO_(3),),("Initial milli-moles",2.5,,1.5,,-,),("Final milli-moles",-,,1,,1.5,):}`
` pH=pK_(a_(1))+log(([HCO_(3)^(-)])/([H_(2)CO_(3)]))=6.173`


Discussion

No Comment Found

Related InterviewSolutions