Question

0.1 mole of `CH_(3)NH_(2)(K_(b)=5xx10^(-4))` is mixed with 0.08 mole of HCl and the solution diluted to one litre. The `H^(+)` ion concentration in the solution will be
A. `1.6xx10^(-11)`
B. `8xx10^(-11)`
C. `5xx10^(-5)`
D. `8xx10^(-2)`

Answer 1

Correct Answer - B
`{:(,CH_(3)NH_(2),+,HCl,rarr,CH_(3)N overset(+)H_(3)Cl^(-),,,),("Initial conc.",0.1,,0.08,,0 "mol" L^(-1),,,),("Conc. in solution after mixing",0.1-0.08 ,,0,,0.08 "mol " L^(-1),,,):}`
`=0.02` (HCl being limiting reactant)
Thus, the final solution is a basic buffer of `CH_(3)NH_(2) and CH_(3)NH_(3)^(+)Cl`
`:. pOH = pK_(b) + log. (["Salt"])/(["Base"])`
i.e., `-log [OH^(-)]=-log K_(b) + log. (["Salt"])/(["Base"])`
or `[OH^(-)]=K_(b) (["Salt"])/(["Base"]) = 5xx10^(-4)xx(0.02)/(0.08) `
`=(5)/(4) xx 10^(-4)M = 1.25 xx 10^(-4)M`
`:. [H^(+)]=(K_(w))/([OH^(-)])=(10^(-14))/(1.25xx10^(-4))=8xx10^(-11)`