pH розчинів багатоосновних кислот

We've been studying acids and bases! We've started to learn how to calculate the pH of both strong and weak acids. The pH of strong acids are easy! They fully dissociate. So, if we calculate their molarity, we have the hydrogen ion concentration. The pH of weak acids is a little more challenging. We have to use the acid dissociation constant. Today, we'll study the pH of polyprotic acids. Remember, they lose multiple hydrogen ions. So, they have two or three dissociation constants. Each dissociation generates more hydrogen ions! How do we calculate their pH? Let's find out!

Carbonic Acid Example

Let's look at carbonic acid. It's a diprotic Arrhenius acid. The first dissociation generates bicarbonate (hydrogen carbonate). In the next dissociation, bicarbonate forms carbonate ions. Its two $pK_{a}$ values are $6.35$ and $10.33$. Let's explore how we calculate the pH.

Let's say we mix a 0.025M solution of carbonic acid. Let's calculate the hydrogen ion concentration from the first dissociation. How would you do it? We'll need to use our equilibrium knowledge. Let's make an ICE table.

First, let's write the balanced chemical equation. It's a simple decomposition reaction.

$H_{2}CO_{3}\rightleftharpoons HCO_{3}^{-}+H^{+}$.

Now we can make our ICE table.

$$\begin{array}{|r|c|c|c|} \hline \text{}&H_{2}CO_{3}&HCO_{3}^{-}&H^{+}\\ \hline \text{Initial}&0.025M&0.0M&0.0M\\ \hline \text{Change}&-x&+x&+x\\ \hline \text{Equilibrium}&0.025-x&+x&+x\\ \hline \end{array}$$

Now we can use that acid dissociation constant to calculate the "x" value. Remember, acid dissociation constants are equivalent to equilibrium constants. But first, let's convert $pK_{a}$ to $K_{a}$

$K_{a1}=10^{-pK_{a}}=10^{-6.35}=4.46\times10^{-7}$

$K_{a1}=\frac{[products]}{[reactants]}\rightarrow 4.46\times10^{-7}=\frac{[x][x]}{0.025-x}$

We have two options for solving. The more manageable method is an approximation. The Ka value is very small. So, we know our reaction stays on the reactant's side. We can assume that carbonic acid's concentration change is negligible. The other option is to use the quadratic formula. But, let's use the approximation.

$10^{-6.35}=\frac{x^{2}}{0.025}\rightarrow x=\sqrt{10^{-6.35}\times 0.025}=\sqrt{1.12\times10^{-8}}=0.00011M$

Our concentration is an order of magnitude less than our initial concentration. So our approximation is valid.

Now, let's calculate the pH based on the first dissociation.

$pH=-log{([H^{+}])}=-log{(0.00011)}=3.9586073$

Typically, we can calculate the pH based on the first dissociation. The reason is that $pK_{a1}>>>pK_{a2}$. So, the bulk of the hydrogen ions will come from the first dissociation. To prove it, let's calculate the pH after the second dissociation.

Again, we'll write a balanced chemical equation.

$HCO_{3}^{-} \rightleftharpoons H^{+} + CO_{3}^{2-}$

And once again, we'll make an ICE table. Remember, this time, our initial hydrogen ion concentration is not zero.

$$\begin{array}{|r|c|c|c|} \hline \text{}&HCO_{3}^{-}&CO_{3}^{2-}&H^{+}\\ \hline \text{Initial}&0.00011M&0.0M&0.00011M\\ \hline \text{Change}&-y&+y&+y\\ \hline \text{Equilibrium}&0.00011-y&+y&0.00011+y\\ \hline \end{array}$$

We calculate the $[H^{+}]$ from our acid dissociation constant.

$K_{a2}=\frac{[products]}{[reactants]}\rightarrow 10^{-10.33}=\frac{[y][0.00011+y]}{0.00011-y}$

Again we should assume that $y<<<0.00011$.

$10^{-10.33}=\frac{[y][0.00011]}{[0.00011]} \rightarrow y=10^{-10.33}=4.7\times10^{-11}$

Let's calculate the new pH.

$pH= -log{(x+y)}=-log{(0.00011M+4.7\times10^{-11})}=3.9586071$

The pH change barely registered! We even went way beyond the correct number of significant figures! If we rounded to the proper decimal place, they're equal. So, we made an excellent assumption! The first dissociation regulates pH.

What's the Point?

Why do we care about polyprotic acids? Later, we'll learn about titrations. They use buffers to help calculate the concentrations of unknown acids and bases. Buffer solutions resist changes to pH. To do so, they must contain an acid and a base. But, we don't want a neutralization reaction. Polyprotic acids create excellent buffers because they form both acids and bases. Consider our carbonic acid example. Carbonic acid produces hydrogen ions. But, it also formed carbonate. If we added a strong acid, it would deplete the carbonate. If we added a strong base, it would consume the hydrogen ions. In fact, you're using the carbonic acid buffer system right now! It's how your blood maintains its pH. So, it's keeping you alive as you read this!

(Video) Polyprotic Acid-Base Equilibria Problems, pH Calculations Given Ka1, Ka2 & Ka3 - Ice Tables

by The Organic Chemistry Tutor

In this video, The Organic Chemistry Tutor shows you four example problems. In each one, you have to calculate the pH of a polyprotic acid. Are you struggling with polyprotic acid problems? If so, this video will help.