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High School Chemistry : Using Base Dissociation Constant (Kb)

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Example Question #1 : Using Base Dissociation Constant (Kb)

A 1M solution of a monoprotic acid has a pH of 4.6. What is the $\small K_{b}$ value for the conjugate base of the acid?

Possible Answers:

$\small 1.58*10^{-5}$

$\small 2.00*10^{-5}$

$\small 4.68*10^{-6}$

$\small 3.98*10^{-10}$

Correct answer:

$\small 1.58*10^{-5}$

Explanation:

In order to find the base dissociation constant for the conjugate base, we can start by finding the acid dissociation constant for the acid. Since a 1M solution of the acid has a pH of 4.6, we can find the proton concentration of the solution.

$pH=-\log[H^+]\rightarrow [H^+]=10^{-pH}$

$\small [H^{+}] = 10^{-4.6} = 2.51*10^{-5}$

Since the acid is monoprotic, we can set the following equilibrium expression equal to its acid dissociation constant.

$\small K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}$

We can see that, since the acid is monoprotic, the concntration of protons will be equal to the concentration of the acid anion. The final concentration of the acid molecule will be equal to the initial concentration, minus the amount of protons formed. Using these values, we can solve for the equilibrium constant for the acid.

$K_a= \frac{[2.51*10^{-5}]^{2}}{[1-2.51*10^{-5}]} = 6.31*10^{-10}$

Now that we have the acid dissociation constant, we can find the conjugate base's dissociation constant by setting the product of the two values equal to the autoionization of water.

$\small K_{w} = K_{a}K_{b}$

$\small 1*10^{-14} = K_{b}(6.31*10^{-10})$

$\small K_{b} = 1.58*10^{-5}$

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High School Chemistry : Using Base Dissociation Constant (Kb)

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