Calculate the [H 3 O + ] and pH of each polyprotic acid solution. b. 0.350 M H 2 C 2 O 4

Hey everyone, welcome back. We're asked to consider a solution of 0.591 molar of carbonic acid and were asked to calculate the concentration of hydrogen ions and the ph of our solution first. If we go ahead and look up our K. S. For our Carbonic acid, we find that our K. A. one Is going to come up to 4.3 Times 10 to the -7. While our K. A. two comes up to 5. Times 10 to the -11. And as we can see when we compare the two K. S. RKA two is extremely small compared to R. K. A. one. So the ph of our solution will only depend on the hydro ni um ions produced from our first equilibrium reaction. So let's go ahead and write that reaction and create our ice chart. We have our carbonic acid and we react this with water and we produce our bicarbonate ion and our hydro ni um ions creating our ice chart. We were told that we started with 0.591 Moeller of our carbonic acid. And we had zero of our products initially. We can go ahead and disregard our water since it is a liquid and our change in our carbonic acid is going to be a minus X. Since we're losing reactant and a plus X in our product side. Since we're gaining products at equilibrium, we have 0.591 minus X. In our react inside and an X. And an X in our product side. Now let's go ahead and use our K. A. To determine the value for X. So we know ourka. was equivalent to 4.3 Times 10 to the -7. And this is going to be equal to our products over our reactant. Now let's go ahead and check if we can disregard our X. In our denominator, we can do so by taking Our 0.591 and dividing this By our K. A. of 4.3 times 10 to the -7. And if we get a value greater than 500 then we can go ahead and disregard our X. And in fact we do get a number greater than 500, we get a value of 1.37 Times 10 to the 5th, Which is greater than 500. So we can go ahead and disregard that X. In our denominator, simplifying this a bit further, we end up with X squared equals 4.3 times 10 to negative seven Times 0. Taking the square root of both sides. We end up with an X. Of 5.04 Times 10 to the negative for Mueller, which is going to be the concentration of our hydro ni um ions. Now let's go ahead and calculate our ph and we can do so by taking the negative log of our concentration of 5.04 Times 10 to the -4. When we do this, we end up with a ph of 3.30. So our final answers here are going to be a ph of 3.30 and a concentration of hydro ni um ions of 5.4 times center negative four Mueller. So I hope this made sense and let us know if you have any questions.

Ch.16 - Acids and Bases

Tro - Chemistry: A Molecular Approach 4th Edition

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Ch.16 - Acids and Bases

Problem 111b

Chapter 16, Problem 111b

Calculate the [H₃O⁺] and pH of each polyprotic acid solution. b. 0.350 M H₂C₂O₄

Verified step by step guidance

Identify that oxalic acid (H2C2O4) is a diprotic acid, meaning it can donate two protons (H+).

Write the dissociation equations for the two steps: \( \text{H}_2\text{C}_2\text{O}_4 \rightleftharpoons \text{H}^+ + \text{HC}_2\text{O}_4^- \) and \( \text{HC}_2\text{O}_4^- \rightleftharpoons \text{H}^+ + \text{C}_2\text{O}_4^{2-} \).

Use the first dissociation constant \( K_{a1} \) to find the concentration of \( \text{H}^+ \) from the first dissociation. Assume \( [\text{H}^+] \approx [\text{HC}_2\text{O}_4^-] \).

Use the second dissociation constant \( K_{a2} \) to find the additional \( \text{H}^+ \) from the second dissociation. This step often contributes less to the \( \text{H}^+ \) concentration.

Calculate the total \( [\text{H}_3\text{O}^+] \) by summing the contributions from both dissociations, then calculate the pH using \( \text{pH} = -\log[\text{H}_3\text{O}^+] \).

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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. Each dissociation occurs in a stepwise manner, with each step having its own dissociation constant (Ka). For example, oxalic acid (H2C2O4) can lose two protons, resulting in H2C2O4 → H+ + HC2O4- and then HC2O4- → H+ + C2O4^2-. Understanding these steps is crucial for calculating the concentrations of H3O+ and pH.

Dissociation Constants (Ka)

The dissociation constant (Ka) quantifies the strength of an acid in solution, indicating how readily it donates protons. For polyprotic acids, there are multiple Ka values corresponding to each dissociation step. The first dissociation typically has a larger Ka than subsequent ones, meaning it is more favorable. These constants are essential for calculating the concentrations of H3O+ ions and determining the pH of the solution.

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+]. For polyprotic acids, the pH is influenced by the contributions from all dissociation steps. To calculate the pH accurately, one must consider the equilibrium concentrations of H3O+ from each dissociation and apply the appropriate Ka values to find the total concentration of H3O+ in the solution.