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  1. Problems
  2. /Phenol has pKa = 10.0 and the anilinium ion has pKa = 4.6…
MediumChemistryHigh School (11–12)Organic Compounds with Functional Groups › AminesNumerical

Phenol has pKa = 10.0 and the anilinium ion has pKa = 4.6…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Phenol has pKa=10.0\text{p}K_a = 10.0pKa​=10.0 and the anilinium ion has pKa=4.6\text{p}K_a = 4.6pKa​=4.6. A phenol couples as its phenoxide ion, and aniline couples as the free amine. Use the ratio of the deprotonated form to the acid form, 10pH−pKa10^{\text{pH} - \text{p}K_a}10pH−pKa​. Find the percentage of phenol that is the phenoxide ion at pH 9.0 and at pH 10.0. Find the percentage of aniline that is the free amine at pH 4.0 and at pH 6.0.

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Need a reminder? The key ideaShowHide

Coupling joins the diazonium ion to the para position of a phenol or an aromatic amine through an azo group, −N=N−-\text{N=N}-−N=N−. The coupling partner reacts in its deprotonated form: the phenoxide ion or the free amine. The fraction in that form is 11+10pKa−pH\dfrac{1}{1 + 10^{\text{p}K_a - \text{pH}}}1+10pKa​−pH1​. For an indicator, [In−][HIn]=10pH−pKa\dfrac{[\text{In}^-]}{[\text{HIn}]} = 10^{\text{pH} - \text{p}K_a}[HIn][In−]​=10pH−pKa​.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. Phenol at pH 9.0. 1001+1010.0−9.0=10011=9.09 %\dfrac{100}{1 + 10^{10.0 - 9.0}} = \dfrac{100}{11} = 9.09\ \%1+1010.0−9.0100​=11100​=9.09 %.

Step 2. Phenol at pH 10.0. 1001+100=50.0 %\dfrac{100}{1 + 10^{0}} = 50.0\ \%1+100100​=50.0 %.

Step 3. Aniline at pH 4.0. 1001+104.6−4.0=1001+3.98=20.1 %\dfrac{100}{1 + 10^{4.6 - 4.0}} = \dfrac{100}{1 + 3.98} = 20.1\ \%1+104.6−4.0100​=1+3.98100​=20.1 %.

Step 4. Aniline at pH 6.0. 1001+104.6−6.0=1001+0.0398=96.2 %\dfrac{100}{1 + 10^{4.6 - 6.0}} = \dfrac{100}{1 + 0.0398} = 96.2\ \%1+104.6−6.0100​=1+0.0398100​=96.2 %.

Answer: 9.09 %9.09\ \%9.09 %, 50.0 %50.0\ \%50.0 %, 20.1 %20.1\ \%20.1 % and 96.2 %96.2\ \%96.2 %.

Why not pH 12 for the phenol? The phenoxide fraction is then almost 100 percent, but the diazonium ion itself is turned into an unreactive diazotate. The best pH is a compromise, at 9 to 10.

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