Calculate the [H 3 O + ] and pH of each polyprotic acid solution. a. 0.125 M H 2 CO 3

Hello everyone today. With the following problem, calculate the concentration of hydro ions and the ph of 0.3 15 molar of sulfurous acid. If we pay attention to the sulfurous acid, we see it has two protons making it a dye pro acid, this means that it can undergo ionization twice. And so for the first ionization reaction, we would have our sulfurous acid in aqueous form reacting of course with a liquid water and this will be a reversible reaction. So we draw the arrows as such. And when one proton dissociates from our sulfurous acid, we get one aqueous proton followed by our other product, we then have our second ionization energy reaction or in our sulfuric sulfurous acid can undergo a ig once again reacting at an aqueous form with liquid water and a reversible reaction to dissociate and to Juan aqueous proton. And this should be aqueous as well. Two hour soul sp which is also aqueous. And if we were to referred to a reference text, we could find the dissociation constant for the first reaction or K A one equal to 1.6 times 10 to the negative second. And then if we look at the second reaction in reference text, the equilibrium value for this association reaction is 6.4 times 10 to the negative eight. Now, since the ionization or acid ionization consonant for the second step is much smaller compared to the first one, the Ph will only depend on the hydro produced from the first equilibrium reaction. So we want to set up next our ice chart and our equilibrium equation. So for our equilibrium equation, we will use our acid ionization constant. Of the first step, we will set that equal to, of course the products, the concentration of the products in the numerator and the concentration of reactants in the denominator. Our products. In this reaction, we had our sulfuric acid and then on them denominator for reactants, we had our hydro ions and then we multiply that by the concentration of our sulfurous acid. And then of course, we said our K a one or acid ionization constant for the first part is equal to 1.6 times 10 to the negative second molar. Now, as for our ice chart, we simply have the reaction for the first dissociation step. We like that with liquid water and reversible reaction to produce our sulfuric acid aqueous. And we also produce hydro of ions that are also aqueous. We have our I for the initial concentration C for the change and then e for our, our final equilibrium values. So the initial concentration of our So first sulfurous acid was 0.3 15 molar. And this was found in the question stem since water is a liquid, it is not included in the and these equilibrium problems and either are solids as for initial concentrations of our products, they were both zero. And since we transfer from the rackets to products, C will be represented by minus X and our rackets will be plus X and then plus X respectively. Our equilibrium is essentially our I plus our C. So for our reactants, we have 3.15 subtracted by X and then we have X and then X for our products. But into our equation, our equilibrium equation, the concentration of our product was said to be X. And since we have, it is said to be represented by X and to determine if X is negligible in the denominator, we have to set up a small equation by taking the concentration of our sulfurous acid in dividing that by our first acid ionization constant, this would be 0.3 15. So divided by 1.6 times 10 to the negative two, which would give us 19.7. And since this is not greater than 500 this is not negligible. So it is not greater than 500. So it is not negligible, meaning that we will include it. So we have our X and the dominator and then we have our 0.3 15 Mueller. This will be subtracted by X. Now we need to finish solving for X. So if we were to do that, if we cross multiply, if we crossed multiplying, we would get the 1.6 times 10 to the negative two multiplied by the X and then the one point, the 0.3 15 minus X that would give us the following value and it should be minus three and we will have X square plus 1.6 times 10 to the negative third, subtracted by 5.04 terms of 10 to the negative three equals zero. Now, to solve for X, we have to bring back the quadratic equation which states that X is equal to negative B in the denominator plus or minus the square root of B squared subtracted by four, multiplied by A multiplied by C. And in the denominator, we have two multiplied by A. In setting this up, our B will be the 1.6 times 10 to negative three. So that will be negative. We have our plus or minus B squared subject that by four, our A was X or just one. And then our C is going to be our 5.04 times 10 to negative three. We divide that by two times A which is just one. And we would get a value that X is equal to both 0.0 634 and negative 0.0794. However, we want to use the positive value. So we can calculate for the Ph. So in calculating for our ph, this will be equal to PH is equal to negative log multiplied by the concentration of hydro ions. So we have the negative log multiplied by the concentration on hydro ions which was 0.0634 giving us a Ph equal to 1.20. And if we look at enter choices and choice d best reflects these values. And with that, we have solved the problem overall, I hope this helped. And until next time.

Ch.16 - Acids and Bases

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Tro 4th Edition

Ch.16 - Acids and Bases

Problem 112a

Chapter 16, Problem 112a

Calculate the [H₃O⁺] and pH of each polyprotic acid solution.
a. 0.125 M H₂CO₃

Verified step by step guidance

Identify that $\text{H}_2\text{CO}_3$ is a diprotic acid, meaning it can donate two protons (H+ ions) in solution.

Write the dissociation equations for $\text{H}_2\text{CO}_3$: $\text{H}_2\text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^-$ and $\text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-}$.

Use the first dissociation constant $K_{a1}$ to find the concentration of $\text{H}^+$ from the first dissociation: $K_{a1} = \frac{[\text{H}^+][\text{HCO}_3^-]}{[\text{H}_2\text{CO}_3]}$.

Assume that the concentration of $\text{H}^+$ from the second dissociation is negligible compared to the first, and calculate $[\text{H}^+]$ using $K_{a1}$.

Calculate the pH using the formula $\text{pH} = -\log[\text{H}^+]$.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Polyprotic Acids

Polyprotic acids are acids that can donate more than one proton (H+) per molecule in a solution. For example, carbonic acid (H2CO3) can lose two protons, resulting in bicarbonate (HCO3-) and carbonate (CO3^2-). Understanding the stepwise dissociation of these acids is crucial for calculating the concentrations of hydronium ions ([H3O+]) and the resulting pH.

Dissociation Constants (Ka)

Each dissociation step of a polyprotic acid has a corresponding acid dissociation constant (Ka), which quantifies the strength of the acid in each step. For H2CO3, the first dissociation (H2CO3 ⇌ H+ + HCO3-) has a different Ka than the second (HCO3- ⇌ H+ + CO3^2-). These constants are essential for calculating the equilibrium concentrations of the species in solution and ultimately determining [H3O+].

pH Calculation

pH is a measure of the acidity of a solution, defined as the negative logarithm of the hydronium ion concentration: pH = -log[H3O+]. To calculate pH for a polyprotic acid, one must first determine the [H3O+] from the dissociation equilibria, considering both dissociation steps. This requires applying the principles of equilibrium and the dissociation constants to find the final concentration of hydronium ions.