- #1

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I've been struggling with this for hours now, but my head doesn't seem to work properly...

## Homework Statement

Aniline, C

_{6}H

_{5}NH

_{2}, is a weak base and reacts with water, forming the phenyl ammonium ion C

_{6}H

_{5}NH

_{3}

^{+}and OH

^{-}.

K

_{b}at 25

^{o}C is [itex]3.2 \cdot 10^{-10}[/itex]

a) What is the equilibrium concentration of aniline at pH=9.00?

b) The solubility of aniline is 3.8g / 100mL of water at 25

^{o}C. Calculate the maximum PH when aniline is dissolved in water at 25

^{o}C.

## Homework Equations

[itex]K_b = \frac{\left[OH^{-} \right]\left[C_6 H_5 NH_3 ^{+} \right]}{\left[C_6 H_5 NH_2 \right]} = 3.2 \cdot 10^{-10}[/itex]

pH + pOH = 14

pOH = -log[OH

^{-}]

## The Attempt at a Solution

I

**think**I did the a) one right:

[itex]pH = 9.00 \, \Longrightarrow \, pOH = 5.00 \, \Longrightarrow \, - \log \left( \left[ OH^{-} \right] \right) = 5.00 \, \Longrightarrow \, \left[ OH^{-} \right] = 10^{-5} = \left[ C_6 H_5 NH_3 ^{+} \right][/itex]

Then;

[itex]\left[ C_6 H_5 NH_2 \right] = \frac{\left[ C_6 H_5 NH_3 ^{+} \right] \cdot \left[ OH^{-} \right]}{K_b} = \frac{(10^{-5})^2}{3.2 \cdot 10^{-10}} = \underline{0.3125 \textrm{ M}}[/itex]

So the equilibrium constant of aniline at pH=9.00 is 0.3125 M.

Is this correct?

Then, I've tried b) for hours... Here are my thoughts:

[itex]n_{\textrm{aniline}} = \frac{m}{M} = \frac{3.8 \textrm{ g}}{93.126 \frac{\textrm{g}}{\textrm{mol}}} = \underline{0.041 \textrm{ mol}}[/itex]

Because it says that 3.8g = 0.041 mol of aniline is dissolved in the water, I assume that this means that there are 0.041 mol of C

_{6}H

_{5}NH

_{3}

^{+}and, equally, 0.041 mol of OH

^{-}in the 100 mL of water.

If so, this means that the concentration of OH

^{-}is 0.41 M.

pOH = -log(0.41) = 0.389 ==> pH = 13.64

My first thought here is that totally neglecting the K

_{b}in this task feels wrong. Also, a pH of >13 for such a weak base also feels very wrong. It also feels wrong to have such high concentrations of ions, when I found in a) that we only had 10

^{-5}of them when we had a 0.3125 M aniline solution.

However, the task said that 3.8g of aniline was dissolved, and 3.8g of aniline is 0.041 mol of aniline, and 0.041 dissolved aniline in 100 mL water, will be 0.41 M of OH

^{-}... At least that is how I interpret the task... So I'm confused right now.

So I guess I screwed it up badly. Where did I go wrong?

Hope you can help me! Thank you very much!!

Daniel