Calculate [CO3^2-] in a 0.010-M solution of CO2 in water (usually written as H2 CO3 ). If all th

step 1 All right, guys, doing problem 110. Chapter 14. It's actually carbonate concentration in a 0 .10

Calculate [CO3^2-] in a 0.010-M solution of CO2 in water...

Calculate $\left[\mathrm{CO}_3^{2-}\right]$ in a $0.010-M$ solution of $\mathrm{CO}_2$ in water (usually written as $\mathrm{H}_2 \mathrm{CO}_3$ ). If all the $\mathrm{CO}_3{ }^{2-}$ in this solution comes from the reaction $$ \mathrm{HCO}_3{ }^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_3{ }^{2-}(a q) $$ what percentage of the $\mathrm{H}^{+}$ions in the solution is a result of the dissociation of $\mathrm{HCO}_3{ }^{-}$? When acid is added to a solution of sodium hydrogen carbonate $\left(\mathrm{NaHCO}_3\right)$, vigorous bubbling occurs. How is this reaction related to the existence of carbonic acid $\left(\mathrm{H}_2 \mathrm{CO}_3\right)$ molecules in aqueous solution?

Calculate $\left[\mathrm{CO}_3^{2-}\right]$ in a $0.010-M$ solution of $\mathrm{CO}_2$ in water (usually written as $\mathrm{H}_2 \mathrm{CO}_3$ ). If all the $\mathrm{CO}_3{ }^{2-}$ in this solution comes from the reaction
$$
\mathrm{HCO}_3{ }^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_3{ }^{2-}(a q)
$$
what percentage of the $\mathrm{H}^{+}$ions in the solution is a result of the dissociation of $\mathrm{HCO}_3{ }^{-}$? When acid is added to a solution of sodium hydrogen carbonate $\left(\mathrm{NaHCO}_3\right)$, vigorous bubbling occurs. How is this reaction related to the existence of carbonic acid $\left(\mathrm{H}_2 \mathrm{CO}_3\right)$ molecules in aqueous solution?

Key Concepts

Le Chatelier’s Principle in Acid–Base Reactions

Le Chatelier’s principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust to partly counteract the change and re-establish equilibrium. In acid–base reactions, this principle explains how the equilibrium shifts when acids or bases are added, leading to phenomena such as gas evolution. For instance, adding acid can shift the equilibrium of bicarbonate dissociation, ultimately causing the release of carbon dioxide gas, which is observed as bubbling.

Carbon Dioxide Hydration and Carbonic Acid Formation

When carbon dioxide dissolves in water, it reacts to form carbonic acid, which then undergoes acid–base equilibria to produce bicarbonate and carbonate ions. This process is an important example of how a gas can equilibrate with its aqueous form, and it plays a significant role in natural buffering systems. Recognizing this interconnectedness between dissolved carbon dioxide and the acid–base reactions of carbonic acid is key to understanding the behavior of the carbonate system in water.

Chemical Equilibrium

Chemical equilibrium refers to the state in which the rate of the forward reaction is equal to the rate of the reverse reaction in a reversible chemical process. This concept is crucial when determining the concentrations of various species in solution, as the equilibrium constant provides a relationship between the concentrations of reactants and products. It underpins the calculation of the concentrations of ions formed from weak acids or bases in aqueous solutions.

Acid?Base Equilibria

Acid–base equilibria describe the dissociation of weak acids and bases in water, leading to an equilibrium between the undissociated molecules and the ions produced. For polyprotic acids such as carbonic acid, which can donate more than one proton, each dissociation step has its own equilibrium constant. Understanding these equilibria is essential to calculate how much of each species (such as bicarbonate and carbonate ions) exists in solution and to assess the contribution of these reactions to the overall ion balance.

Transcript

00:01 All right, guys, doing problem 110.

00:02 Chapter 14.

00:03 It's actually carbonate concentration in a 0 .10 molar solution of co2 in water.

00:08 They're usually written as h2 -c -o -3.

00:11 Well, the co3 -2 -mise solution comes from the reaction.

00:16 Action, h -c -o -3 -minus yields h -plus -c -o -3 -1.

00:22 A percentage of the h -plus ions is a result dissociation of h -c -o -3 mice.

00:28 So, carbonic acid will dissociate.

00:32 Into h plus and hco3 minus and then hco3 minus, but we'll further associate into h plus and co3 and 2 minus.

00:44 So first they want to figure what percentage of hydrogen ions are due this.

00:48 So we're going to plug everything into our equation, our ka equation.

00:54 So the ka for carbonic acid, that's 5 times 10 to the negative 7.

01:27 And that's going to equal x squared, which is going to be the product of h -plus concentration and h -co3 -mine's concentration over the concentration of our acid.

01:39 Let's assume we have less than 5 % association.

01:42 So then that's going to be x squared is equal 2 .2 times 10 to the negative 4th.

01:59 Sorry, that is 5 times 10 to the 8, 10 to negative 8.

02:11 That means x is 2 .2 times 10 to the e .5th molar.

02:22 So that's the, and we can see that several orders of magnitude smaller than 0 .10, so that's less than 5 % ionization.

02:31 So the our h1, so our first h plus concentration, that's going to be equal to this.

02:48 That's also the concentration of our hydrogen carbonate, or rather our bicarbonate.

02:56 So next we're going to look at pca2 and see how much protons we get from that.

03:02 So 10 to the negative.

03:04 So 10 to.

03:12 So this is going to be 4 .8.

03:17 Times 10 to the negative 11 divided by 2 .2 times 10 to the negative 4th.

03:30 Let's assume less than 5 % ionization.

03:40 X squared is equal to 1 .1 times 10 to the negative 14.

03:51 And that means x is going to be 1 % 1 times 10.

04:09 To the negative 7th molar.

04:12 We can see that this is several or as magnitude less than 2 .2 times 10 to negative 4.

04:17 So that's going to be, that's going to be less than 5 % in the initialization.

04:29 So this is going to be equal to h plus 2...

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