Обчислення йонних добутків при йонізації багатоосновних кислот

We've been studying polyprotic acids. They have multiple ionizations. So, they have successive ionization constants. Before, we investigated their ionization constants. We saw that for each dissociation, the constant decreased. But, the first dissociation determines the pH. Remember, for acids and bases we can also use the dissociation constants.

Today, we'll learn to use successive dissociations to calculate product concentrations. Thankfully, weak acids and bases are just equilibrium problems. So, we can use our equilibrium constant calculations.

Carbonic Acid Example

carbon dioxide and water react to form carbonic acid. Carbonic acid is a weak Arrhenius acid. It's responsible for the buffer capacity of the ocean and your blood. Its conjugate is the bicarbonate anion, which is amphoteric.

In soda, the partial pressure of carbon dioxide is 2.5 atmospheres! Under typical pressures, carbon dioxide and carbonic acid are in equilibrium. But, at 2.5 atm, we have a saturated solution. Consider Le' Chatelier's principle. We have a lot of reactant. So, we'll push our equilibrium towards the products. We've shifted the balance so much that we've changed the dissociation constants. Let's explore the pH of soda!

At 2.5atm of carbon dioxide, the molarity of carbonic acid is $1.43\times 10^{-4}$. Find the concentration of bicarbonate and carbonate ions. Find the pH of the final solution. The new $pKa$ values are 3.60 and 10.25.

First, we need to write our balanced chemical equations.

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

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

Next, we need to make an ICE table for the first dissociation.

$$\begin{array}{|r|c|c|c|} \hline \text{}&H_{2}CO_{3}&HCO_{3}^{-}&H^{+}\\ \hline \text{Initial}&1.43\times 10^{-4}&0.0M&0.0M\\ \hline \text{Change}&-x&+x&+x\\ \hline \text{Equilibrium}&1.43\times 10^{-4}-x&+x&+x\\ \hline \end{array}$$

Now, we can solve for the concentration change, "x."

$K_{a1}=\frac{[products]}{[reactants]}\rightarrow 10^{-3.60}=\frac{[x][x]}{1.43\times 10^{-4}-x}$

We should use the approximation method. That removes the "-x" in the denominator.

$10^{-3.60}=\frac{x^{2}}{1.43\times 10^{-4}}\rightarrow x=\sqrt{10^{-3.60}\times 1.43\times 10^{-4}}=\sqrt{3.59\times10^{-8}}=1.90\times10^{-4}$

It may seem like our approximation is invalid. We would use up all our carbonic acid. But, we have so much carbon dioxide that we can assume excess carbonic acid. The carbon dioxide will continually generate carbonic acid. So, what stops our reaction? The equilibrium between bicarbonate and carbonate.

So, $[H^{+}]=[HCO_{3}^{-}]=1.4\times10^{-4}$. And, $pH=-log{(1.90\times10^{-4})}=3.72$.

Now, we can use $pK_{a2}$ to solve for $[CO_{3}^{2-}]$.

$K_{a2}=\frac{[products]}{[reactants]}\rightarrow 10^{-10.25}=\frac{[H^{+}][CO_{3}^{2-}]}{[HCO_{3}^{-}]}\rightarrow [CO_{3}^{2-}]=10^{-10.25}=5.62\times10^{-11}$

So, what are the final concentrations of the products and the pH?

$[HCO_{3}^{-}]=1.90\times10^{-4}M$

$[CO_{3}^{2-}]=5.62\times10^{-11}M$

$[H^{+}]=1.90\times10^{-4}M \rightarrow pH=3.72$

The second dissociation does not affect pH or bicarbonate ion concentration. That's because $pK_{a1}>>>pK_{a2}$. So, the second dissociation is minuscule! We only needed one ICE table. Then we can use the ionization constant to calculate the carbonate concentration.

Chemistry in the World: Cavities

In our carbonic acid example problem, we saw that soda's pH was 3.72! That crazy acidic. The enamel is the outside layer of a tooth. It begins to dissolve at a pH below 5.50. As the enamel dissolves, it exposes dentin. Dentin is a hardened calcium-based mineral. It provides strength and structure to your teeth. Dentin begins to dissolve at a pH of 6.50. Even worse, it's porous! So it has a very high surface area. Over time, the acidity erodes the dentin. What's the result? You get cavities! The nerves at the center of your teeth become exposed. They are unbelievably sensitive to temperature and pH. Now, when you drink a cold drink, you experience shooting pain. Unfortunately, at this point, it's time to see the dentist.

(Video) pH calculation for a polyprotic acid

Diego Troya

In this video, Diego walks you through the dissociation of carbonic acid. He shows you each ICE table. So, you'll see how to calculate the products for each dissociation. This video will help you understand how to calculate the products of polyprotic acids.