Calculate the [H 3 O + ] and pH of each polyprotic acid solution. a. 0.350 M H 3 PO 4

Hey everyone, we're asked, what is the ph and concentration of our protons of a 0.24 Mueller polyp protic acid First looking at R. K. S, we're going to use our first K. A. Since it is much larger than our 2nd and 3rd K. Now let's go ahead and write out our reaction. So we have our polyp protic acid which will symbolize as H3 a. And we react this with water. We then form the conjugate base of our product acid, which will be H two a minus plus our hydro ni um ions. We can then go ahead and create our ice chart. Initially, we were told that we had 0.24 Mueller of our polyp protic 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 is going to be a minus X. In our reacting side. Since we're losing reactant and a plus X in our product side. Since we're gaining products At Equilibrium, we're going to have 0.24 Moeller -X. For our react inside an X. And an X in our product side. For our K. A. one. We know that this is going to be our products over our reactant and this is going to be equivalent to our K. A. One of 6.3 times 10 to the negative third First. Let's go ahead and check if we can disregard our X. And we can do so by taking our 0.24 And dividing it by our K. A. one of 6.3 times 10 to the negative third. If we get a value Less than 500 then we cannot disregard it. And in this case we got a value of 37 Which is in fact less than 500. So we have to keep our X solving for R. X. We can further simplify this and we end up with X squared over 0.24 minus X equals 6. times 10 to the negative third. We can go ahead and multiply both sides by 0.24 -1. We then get X squared equals 6.3 times 10 to the negative third Times 0.24 -X. And since we do need to use our quadratic equation we can rewrite this as X squared plus 6.3 times 10 to the negative third. X minus 6.3 times 10 to the negative 3rd times 0.24. Now that we have it in this form, let's go ahead and use our quadratic equation to solve for X. Using our quadratic equation, we end up with X as 0.03586. So this means our concentration of hydro ni um ions is 0.3586, Mohler. Now in order to find our ph we can then take the negative log of our concentration of hydro knee um ions which is 0.3586, And this will get us to a ph of one 0.45. So our final answers here are going to be a ph of 1. and a concentration of hydro ni um ions of 0.36 Mohler, since we have two significant figures, so I hope this made sense and let us know if you have any questions.

Ch.16 - Acids and Bases

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Tro 4th Edition

Ch.16 - Acids and Bases

Problem 111a

Chapter 16, Problem 111a

Calculate the [H₃O⁺] and pH of each polyprotic acid solution. a. 0.350 M H₃PO₄

Verified step by step guidance

Recognize that H3PO4 (phosphoric acid) is a polyprotic acid, meaning it can donate more than one proton (H+). It dissociates in steps: H3PO4 -> H2PO4^- + H+, H2PO4^- -> HPO4^2- + H+, and HPO4^2- -> PO4^3- + H+.

The first dissociation of H3PO4 is the most significant in determining the [H3O+] and pH because it has the largest equilibrium constant (Ka1). Write the equilibrium expression for the first dissociation: Ka1 = [H2PO4^-][H3O+]/[H3PO4].

Assume that the initial concentration of H3PO4 is 0.350 M and that it dissociates to form x M of H3O+ and H2PO4^-. The equilibrium concentrations will be [H3PO4] = 0.350 - x, [H2PO4^-] = x, and [H3O+] = x.

Use the known value of Ka1 for H3PO4 (approximately 7.5 x 10^-3) to solve for x: Ka1 = x^2 / (0.350 - x). Since Ka1 is relatively small, you can assume x << 0.350 and simplify the equation to Ka1 ≈ x^2 / 0.350.

Once you have solved for x, which represents [H3O+], calculate the pH using the formula pH = -log[H3O+]. This will give you the pH of the solution.>

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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 step has its own equilibrium constant, known as Ka, which decreases with each successive proton donation. Understanding the behavior of polyprotic acids is crucial for calculating the concentrations of hydronium ions ([H3O+]) and the resulting pH in solutions.

Dissociation Constants (Ka)

The dissociation constant (Ka) quantifies the strength of an acid in solution, indicating how well it donates protons. For polyprotic acids, there are multiple Ka values corresponding to each dissociation step. These constants are essential for determining the concentration of [H3O+] in a solution, as they help predict the extent of dissociation for each proton donation.

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 polyprotic acids, one must first determine the [H3O+] from the dissociation equilibria, considering the contributions from all dissociation steps. This calculation is vital for understanding the acid's strength and its effect on the solution's properties.