At $\pu{25°C}$, a $\pu{0.100 M}$ aqueous solution of $\ce{C_6H_5NH_2}$ has $\mathrm{pH}=8.80$. Find the $K_{\mathrm{b}}$.

I figured I would have to use an ICE table, but I am not too sure how to set it up.

I can find the equation, which is $$\ce{C6H5NH2 -> C6H5NH3^+ +OH-}$$ but I don't think it helps. I only have an elementary knowledge of acid/bases, so I am hoping for a simple answer.


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    $\begingroup$ I've never used $K_{b}$ before, but it sounds like $K_{a}$. So what you could do is approximate the numerator in the $K_{b}$ equation to $[OH^{-}]^{2}$. Then find $[OH^{-}]$ by calculating $[H^{+}]$ and using $K_{w}$. Finally, use the data you've calculated (and been given) to calculate $K_{b}$. $\endgroup$ – user60221 Apr 24 '18 at 7:37
  • $\begingroup$ @KianStevens $K_b$ is $(1.0\cdot10^{-14})/K_a$, and I'll try that thanks $\endgroup$ – suomynonA Apr 24 '18 at 8:00
  • $\begingroup$ Your given equation is incomplete. It needs a water molecule in left hand side. $\endgroup$ – Mathew Mahindaratne Apr 24 '18 at 15:31
  • $\begingroup$ have a read of this answer and the link. chemistry.stackexchange.com/questions/95681/… It explains how to do these type of questions. $\endgroup$ – porphyrin Apr 24 '18 at 16:10

$\ce{C6H5NH2}$ (aniline) is a weak base, and therefore, in its aqueous solution, it takes a proton from water and forms $\ce{C6H5NH3^+}$ and $\ce{OH-}$, which are correct as you have written.

Suppose the extent of protonation is $\alpha$. After taking up the proton, $[\ce{C6H5NH2}] = c (1 - \alpha)$ and $[\ce{C6H5NH3^+}]= c \alpha = [\ce{OH-}]$.

So, we can write the following, from the above mentioned reaction:

$$ K_{\mathrm{b}} = \frac{[\ce{C6H5NH3^+}][\ce{OH-}]}{[\ce{C6H5NH2}]} = \frac{c^{2}\alpha^{2}}{c(1 - \alpha)} \approx c \alpha^{2} $$

As the degree of protonation is very less than 1 ($\alpha << 1$), we get $\alpha = \sqrt{K_{\mathrm{b}} / c}$. Now,

\begin{align*} \mathrm{p}\ce{OH} &= -\log[\ce{OH-}] \\ & = -\log(c\alpha)=-\log(c\sqrt{K_{\mathrm{b}}/ c})=-\log(\sqrt{cK_{\mathrm{b}}}) \\ &= \frac{\mathrm{p}K_{\mathrm{b}}}{2} - \frac{\log c}{2} \end{align*}

(as $-\log K_{\mathrm{b}} = \mathrm{p}K_{\mathrm{b}}$).

In the question, $\mathrm{pH} = 8.8$, so, $\mathrm{pOH} = 14 - 8.8 = 5.2$, and $c = 0.1$. Putting these values in the equation we will get

$$ \mathrm{p}K_{\mathrm{b}} = 2.(5.2 + \frac{\log(0.1)}{2}) = 9.4 \implies K_{\mathrm{b}} = 10^{-9.4} = 3.98 \times 10^{-10} $$

Thus we get the $K_{\mathrm{b}}$ of aniline as $3.98 \times 10^{-10}$.

Note that, if you have any aqueous solution consisting of only weak base, you can apply the direct formula as $\mathrm{pOH} = \frac{\mathrm{p}K_{\mathrm{b}}-\log c}{2}$, and easily get the $K_{\mathrm{b}}$.

  • $\begingroup$ Thanks! I'm not a huge fan of memorizing formulas but I'm sure I can derive it pretty quickly now $\endgroup$ – suomynonA Apr 24 '18 at 23:29

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