Question

A 0.10 M imidazole solution has a pH of 6.6. To the nearest hundredth of a unit, what fraction of the molecules are in the neutral (imidazole) form? (The pKa of the imidazolium ion is 6.0.)

Answer 1

Answer : The fraction of the molecules in the neutral (imidazole) form are, 0.799

Explanation : Given,

pH = 6.6

$p_{K_a}=6.0$

Using Henderson Hesselbach equation :

$pH=pK_a+\log \frac{[Salt]}{[Acid]}$

$pH=pK_a+\log \frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}$

Now put all the given values in this expression, we get:

$6.6=6.0+\log \frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}$

$\frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}=10^{6.6-6.0}$

$\frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}=10^{0.6}$

$\frac{[\text{Imidazole}]}{[\text{Imidazolium ion}]}=3.98$

$[\text{Imidazole}]=3.98[\text{Imidazolium ion}]$     ...........(1)

Now we have to determine the fraction of the molecules are in the neutral (imidazole) form.

Fraction of neutral imidazole = $\frac{[\text{Imidazole}]}{[\text{Imidazole}]+[\text{Imidazolium ion}]}$

Now put the expression 1 in this expression, we get:

Fraction of neutral imidazole = $\frac{3.98[\text{Imidazolium ion}]}{3.98[\text{Imidazolium ion}]+[\text{Imidazolium ion}]}$

Fraction of neutral imidazole = $\frac{3.98[\text{Imidazolium ion}]}{4.98[\text{Imidazolium ion}]}$

Fraction of neutral imidazole = $\frac{3.98}{4.98}$

Fraction of neutral imidazole = 0.799

Thus, the fraction of the molecules in the neutral (imidazole) form are, 0.799