Question

When `100 mL` of `0.1 M KNO_(3)` and `400 mL` of `0.2 MHCl` and `500 mL` of `0.3 M H_(2) SO_(4)` are mixed, then in the resulting solution
A. The molarity of `K^(o+) = 0.01 M`
B. The molarity of `SO_(4)^(2-) = 0.15 M`
C. The molarity of `H^(o+) = 0.38 M`
D. The molarity of `NO_(3)^(ɵ) = 0.08` and `Cl^(ɵ) = 0.01 M`

Answer 1

Correct Answer - A::B::C
`underset(({:(100 xx 0.1),(=10"mol"):}))(KNO_(3)) + underset(({:(400xx0.2),(=80"mmol"):}))(HCl) + underset(({:(500xx0.3),(=150"mmol"):}))(H_(2)SO_(4))`
Total volume `= 100 + 400 + 500 = 1000 mL`
`M_(K^(o+)) = (10)/(1000) = 0.01 M`
`M_(SO_(4)^(-2)) = (150)/(1000) = 0.15 M`
`M_(H^(o+)) = (80 + 2 xx 150)/(1000) = 0.38 M`
`M_(NO_(3)^(ɵ)) = (10)/(1000) = 0.01 M`
`M_(Cl^(ɵ)) = (80)/(1000) = 0.08 M`