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.