Question

show work please

1) The structure of the compound imidazole is shown below, with its ionization reaction and pka. a) At PH 7.5, which form of imidazole do you expect to predominate? (1 pt) b) What percentage of imidazole is charged at pH 7.5? (3 pt)

Answer 1

a)

Note that the one that produces H+ can be denoted as the acid and its conjugate base that forms can be denoted as the base.

pka H+ + HN [acid] [base]

Now, we can write the expression of dissociation constant Ka as follows

$K_a = \frac{[H^+][base]}{[acid]}$

Now, we know that pKa of the acid is defined as the negative logarithm of Ka.

Hence, taking negative logarithm on both sides, we get

$-\log K_a = -\log \left (\frac{[H^+][base]}{[acid]} \right ) \\ \\ \Rightarrow pK_a = -\log[H^+] - \log \left ( \frac{[base]}{[acid]} \right )$

Note that negative logarithm of H+ concentration is the pH of the solution. Hence,

$pK_a = pH - \log \left ( \frac{[base]}{[acid]} \right )$

For us, pH = 7.5 and pKa = 7.0

Hence, putting in the values,

$pK_a = pH - \log \left ( \frac{[base]}{[acid]} \right ) \\ \\ \Rightarrow 7.0 = 7.5 - \log\left ( \frac{[base]}{[acid]} \right ) \\ \\ \Rightarrow \log \frac{[base]}{[acid]} = 7.5-7.0 = 0.5 \\ \\ \Rightarrow \frac{[base]}{[acid]} = 10^{0.5} \approx 3.16$

Hence, clearly, the unionized basic form of the imidazole will predominate at pH 7.5 as its concentration is almost 3 times more than the concentration of the acidic imidazolium cation.

Hence, the form on the product side will predominate.

(b)

The charged imidazolium cation is the acid.

The uncharged imidazole is the base.

Hence, we can write

$[acid] + [base]= 100\%$.

Now, we already know that

$\frac{[base]}{[acid]} = 3.16$

Hence, the percentage of charged imidazole i.e. acid can be calculated by substitution as follows

$[acid] + [base]= 100\% \\ \Rightarrow [acid] + 3.16 \times [acid] = 100\% \\ \Rightarrow 4.16\times [acid] 100 \% \\ \Rightarrow [acid] = \frac{100\%}{4.16} \approx 24.0\%$

Hence, 24.0 % of imidazole is charged at a pH of 7.5.