0.1 mole of `CH_3NH_2` `(K_b=5xx10^(-4))` is mixed with 0.08 mole of HCl and diluted to 10 litre, what will be the `OH^-` concentration in the solution ?

0.1 mole of CH_(3)CH (K_(b) = 5 xx 10^(-4)) is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H^(+) concentration in the solution

0.1 mole of CH_(3)NH_(2) (K_(b)=5xx10^(-4)) is mixed with 0.08 mole of HCl and diluted to one litre. The [H^(+)] in solution is

0.1mol of CH_(3)NH_(2) (K_(b) = 5 xx 10^(-4)) is mixed with 0.08mol of HCI and diluted to 1L . Which statement is correct?

0.1mol of RNH_(2)(K_(b) = 5 xx 10^(-5)) is mixed with 0.08mol of HC1 and diluted to 1L . Calculate the [H^(o+)] in the solution.

HH_(3)NH_(2)(0.1"mole",K_(b)=5xx10^(-4)) is added to 0.08 moles of HCI and the solution is diluted to one litre. The resulting hydrogen ion concentration is