pH спряжених розчинів

We've been studying acid and bases. We know qualitatively that weak acids and bases produce strong conjugates. For example, we studied the acid-base properties of salts. Or if we neutralize a weak acid/base with a strong base/acid, we generate a conjugate. But, so far, we haven't calculated the pH of conjugates. Let's explore how to do it!

The Relationship Between an Acid and its Conjugate Base

Let's say that we had a generic weak acid dissociation.

$HA_{(s)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O_{(aq)}^{+} + A_{(aq)}^{-}$

What happens when the conjugate base reacts with water? It generates hydroxide ions. This is a form of the hydrolysis reaction.

$A^{-}_{(aq)} + H_{2}O_{(l)} \rightleftharpoons OH_{(aq)}^{-} + HA_{(aq)}^{-}$

Weak acids and bases are in equilibrium. So, we can write two ionization constants expressions. The first equation is the acid's dissociation. The second equation is the base's dissociation. Remember, water is a pure liquid. So, we exclude it from the constant.

$K_{a}=\frac{[H_{3}O^{+}][A^{-}]}{[HA]}$

$K_{b}=\frac{[HA]}{[OH^{-}][A^{-}]}$

Now let's multiply the expressions.

$K_{a}\times K_{b}= \frac{[H_{3}O^{+}]\cancel{[A^{-}]}}{\cancel{[HA]}} \times \frac{\cancel{[HA]}}{[OH^{-}]\cancel{[A^{-}]}}=[H_{3}O^{+}][OH^{-}]=K_{w}$

So, what's the product of the acid and conjugate base ionization constants? It's the water ionization constant!

Now let's take the logarithm of both sides.

$\log{(K_{w})} = \log{(K_{a}\times K_{b})}\rightarrow pK_{w}=14=pK_{a} + pK_{b}$.

So, now if we have acid dissociation constant ($pK_{a}$) we can solve for the base dissociation constant ($pK_{b}$).

Finding the pH of a Conjugate

We can use our derived relationship to calculate the pH of a conjugate.

Conjugate Base Example

Let's consider an example. Let's say we want to find the pH of 0.02 molar solution of sodium acetate ($NaCH_{3}COO$). Acetate is the conjugate of acetic acid. It's a weak Arrhenius acid. Acetic acid's $pK_{a}=4.75$.

Let's start by writing the balanced chemical equation.

$CH_{3}COO^{-} + H_{2}O \rightleftharpoons CH_{3}COOH + OH^{-}$

Let's find $K_{b}$.

$pK_{w}=14=pK_{a} + pK_{b} \rightarrow pK_{b}=14.00-4.75=9.25 \rightarrow K_{b}=10^{-9.25}$.

Now we can solve for the hydroxide ion concentration.

$K_{b}=10^{-9.25}=\frac{[CH_{3}COOH][OH^{-}]}{[CH_{3}COO^{-}]}$

Here, it's easiest to perform an approximation. In our chemical equation, the mole ratio is 1:1. So, each mole of acetate ion generates one mole of both hydroxide and acetic acid. Therefore, $[CH_{3}COOH]=[OH^{-}]$.

$k_{b}=10^{-9.25}=\frac{[CH_{3}COOH][OH^{-}]}{[CH_{3}COO^{-}]} \rightarrow 10^{-9.25}=\frac{[OH^{-}]^{2}}{[CH_{3}COO^{-}]} \rightarrow [OH^{-}]=\sqrt{10^{-9.25}\times 0.02}= 3.35\times 10^{-6}$

We have the $[OH^{-}]$! Now, we need to solve for $[H^{+}]$. We can use either $K_{w}$ or pOH.

$K_{w}=[H^{+}][OH^{-}]\rightarrow [H^{+}]=\frac{10^{-14}}{3.35\times10^{-6}}= 2.98\times 10^{-9}$

Finally, we can solve for the pH.

$pH=-log{[H^{+}]}=-log{(2.98\times 10^{-9})}=8.53$

So, we see that our solution became slightly basic. That makes sense because we added a small concentration of a strong base!

Conjugate Acid Example

What happens when we add a conjugate acid? Let's say we create a solution of 0.1 molar methyl-ammonium chloride. Methyl-ammonium is the conjugate of methylamine. Remember, amines are Lewis bases because of nitrogen's lone pair of electrons. The $pK_{b}$ of methylamine is 3.34. Let's calculate the pH!

Again, let's write a balanced chemical equation.

$CH_{3}NH_{3}^{+} + H_{2}O \rightleftharpoons H_{3}O^{+} + CH_{3}NH_{2}$

Now, we can solve for $K_{a}$.

$pK_{w}=14=pK_{a} + pK_{b} \rightarrow pK_{a}=14.00-3.34=10.66 \rightarrow K_{a}=10^{-10.66}$.

Now we can solve for $[H^{+}]$. Again, it's best to use an approximation.

$[H^{+}]= \sqrt{[CH_{3}NH_{3}^{+}]\times K_{a}}=1.48\times 10^{-6}$

$pH =-log{([H^{+}])}=5.83$

Again, we can confirm our answer qualitatively. We added a small amount of acid and our pH dropped! Everything checks-out!

(Video) 16.6 Acidity and Basicity of Salts | General Chemistry

by Chad's Prep

In this section of the video, Chad explains how to calculate the pH of salt solutions. That's the same as calculating the pH of conjugates! He'll show you an example of a conjugate base and a conjugate acid. This video is a great resource for learning how to do these calculations.