Question

`2.5 ` mL of `(2)/(5)` M weak monoacidic base `(K_(b)=1xx10^(-12) "at" 25^(@)C)` is titrated with `(2)/(15)` M HCl in water at `25^(@)` C. The concentration of `H^(+)` at equivalence point is `(K_(w) = 1 xx 10^(-14) "at" 25^(@)C)`
A. `3.7xx10^(-13)M`
B. `3.2xx10^(-7) M`
C. `3.2xx10^(-2)M`
D. `2.7xx10^(-2)M`

Answer 1

Correct Answer - C
`{:(N_(1)V_(1),=,N_(2)V_(2)),("(Base)",,"(Acid)"):}`
`2.5xx(2)/(5)=(2)/(15)=(2)/(15)xxV_(2) "or" V_(2)=(15)/(2) mL = 7.5 mL `
`BOH+HClrarrBCl+H_(2)O`
2.5 mL of 2 M base contain base
`=2.5xx(2)/(5) = 1` mmol
`:. ` Salt BCl formed = 1 mmol
Volume of solution = 2.5 mL + 7.5 mL = 10 mL
`:.` Conc. of salt [BCl] in the solution
`=(1)/(10)M=0.1M`
For salt of weak base and strong acid
`[H^(+)]=sqrt((K_(w)C)/(K_(b)))=sqrt((10^(-14)xx0.1)/(10^(-12)))=3.2xx10^(-2)M`