H+ Calculation for 0.220 M Aniline Solution with Kb = 7.4e-10

[parwana]

Calculate [H+]?

Calculate [ H + ] of a 0.220 M solution of Aniline C6H5NH2 Kb = 7.4e-10

 

[Sirus]

Consider that for a reaction
HA_{(aq)}\rightleftharpoons~H^{+}_{(aq)}~+~A^{-}_{(aq)}
we can use
K_{a}=\frac{[H^{+}_{(aq)}][A^{-}_{(aq)}]}{[HA_{(aq)}]}
How can you find K_{a} when given K_{b}?

 

[chem_tr]

Sirus is right; an alternative point of view may be using dissociative approach. In this, you start with 0.220 M of aniline, but only \displaystyle x of it is ionized to give some \displaystyle H^+. We know the equilibrium constant of this reaction, i.e., \displaystyle \frac {10^{-14}}{K_b}.

\displaystyle \underbrace{C_6H_5NH_2_{(aq)}}_{0.220-x} \leftrightharpoons \underbrace{C_6H_5NH^-_{(aq)}}_{x}+\underbrace{H^+_{(aq)}}_{x}

Now it is better for you to consider the magnitude of \displaystyle \frac {10^{-14}}{K_b}; for the sake of simplification, you may omit the \displaystyle x in \displaystyle 0.220-x, as we may omit values generally less than 5%. Of course, you may also want to solve the quadratic equation to find the exact answer, but believe me this is not very necessary.

What you'll do next is to find the \displaystyle -\log x.

 

[ShawnD]

Sirus is right; an alternative point of view may be using dissociative approach. In this, you start with 0.220 M of aniline, but only \displaystyle x of it is ionized to give some \displaystyle H^+. We know the equilibrium constant of this reaction, i.e., \displaystyle \frac {10^{-14}}{K_b}.

\displaystyle \underbrace{C_6H_5NH_2_{(aq)}}_{0.220-x} \leftrightharpoons \underbrace{C_6H_5NH^-_{(aq)}}_{x}+\underbrace{H^+_{(aq)}}_{x}

Aniline is a base, not an acid. Hydrogen will stick to the lone pair on the nitrogen so the reaction is like this:

C_6H_5NH_2 + H_2O \rightarrow C_6H_5NH_3^+ + OH^-

 

[Sirus]

In that case, for a reaction
B_{(aq)}~+~H_{2}O_{(l)}\rightleftharpoons~BH^{+}_{(aq)}~+~OH^{-}_{(aq)}
we can use
K_{b}=\frac{[BH^{+}_{(aq)}][OH^{-}_{(aq)}]}{[B_{(aq)}]}

 

[chem_tr]

Dear ShawnD, you are right about aqueous reactions, but don't forget that there are very powerful bases for using in non-aqueous phases like sodium hydride, lithium diisopropylamide, etc, which can take a proton to give anilinide anion.

In most cases, your reaction is sufficient, and same things may be said for Sirus' last post.