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

0.1 mole of CH_(3)NH_(2)(K_(b)=5xx10^(-4)) is mixed with 0.08 mole of

1.	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
Answer» `1.6xx10^(-11)` `8xx10^(-11)` `5XX10^(-5)` `8xx10^(-2)` Solution :`{:(,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 abasic 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)`