Calculate the [H 3 O + ] and pH of each polyprotic acid solution. b. 0.125 M H 3 C 6 H 5 O 7

Welcome back everyone. What is the Ph and the concentration of hydro num for the poly protic acid solution provided below. We're given 0.15 molar C seven H 603 with two K A values. Meaning it is a Dipro acid. Essentially, what we want to understand is that assets, if they are weak assets and we are given their low K values, they will in water to produce hydro num. So if this is a dip protic acid, we can just use a general form of H two A, right? We understand that there are two acidic hydrogens within this acid and A is just a conjugate base. So essentially what we're going to do is just consider the first ionization step where we are losing the first hydrogen to what are to produce hydro, we are producing H A minus and hydro. And we essentially have to recall that whenever we want to identify the concentration of hydro and ph, we are essentially constructing an IC table in terms of molarity here. So initially, we have a concentration of 0.15 molar and we can just say that we have zero, approximately zero for hydro, the change would be negative X and positive X for the product side. And at equilibrium that would be 0.15 minus XX and X. And we know that this is the first ionization step. So K A one would be equal to X square divided by 0.15 minus X. We're simply taking the products multiplying their concentrations and dividing by the concentration of the reactant equilibrium. And our essential, we want to see whether or not we have to solve the quadratic equation to do that. We are going to take the initial concentration and divide it by K A one. The initial concentration is 0.15. And we're going to divide it by 8.3 multiplied by its sense. The power of negative fifth. What we notice is that this is much, much greater than 1000 meaning negative X in the denominator is negligible. And if we solve this equation, we can essentially simplify it as ke one is approximately equal to X squared divided by 0.15. So our X is simply square root off the product between the initial concentration and K A one which is the concentration of hydro at equilibrium. We're going to neglect the second ionization step, right? Because as we understand this is a very weak acid and we can also neglect the value of X in the denominator. The. Therefore, the second ionization step will not matter because K A two is also much lower than K A one. So if we solve for X, we get a value of, we need to use two significant figures. And that'd be 3.5 multiplied by sense, the power of negative third molar. Now this would be the concentration of hydro and now our ph is negative log of the concentration of hydro by definition. So we're just going to take the previously calculated value in order to avoid the propagation of air. We're just going to substitute square root of the product between the concentration and K A one. And this gives us the P value of 2.45. So we have our concentration of hydro and also the ph of the solution which is 2.45. So let's label our answers. That would be it. Thank you for watching.

Ch.16 - Acids and Bases

Tro - Chemistry: A Molecular Approach 4th Edition

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Ch.16 - Acids and Bases

Problem 112b

Chapter 16, Problem 112b

Calculate the [H₃O⁺] and pH of each polyprotic acid solution.
b. 0.125 M H₃C₆H₅O₇

Verified step by step guidance

Identify that citric acid (H_3C_6H_5O_7) is a triprotic acid, meaning it can donate three protons (H^+).

Write the dissociation reactions for each step: \( \text{H}_3\text{C}_6\text{H}_5\text{O}_7 \rightleftharpoons \text{H}^+ + \text{H}_2\text{C}_6\text{H}_5\text{O}_7^- \), \( \text{H}_2\text{C}_6\text{H}_5\text{O}_7^- \rightleftharpoons \text{H}^+ + \text{HC}_6\text{H}_5\text{O}_7^{2-} \), and \( \text{HC}_6\text{H}_5\text{O}_7^{2-} \rightleftharpoons \text{H}^+ + \text{C}_6\text{H}_5\text{O}_7^{3-} \).

Use the first dissociation constant (\(K_{a1}\)) to calculate the initial [H_3O^+] concentration, assuming the first dissociation is the most significant.

Set up an ICE table (Initial, Change, Equilibrium) for the first dissociation to find the equilibrium concentrations of all species.

Calculate the pH using the formula \( \text{pH} = -\log[\text{H}_3\text{O}^+] \) once the [H_3O^+] is determined from the first dissociation.

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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 concentration 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 calculating the concentrations of ions in solution, particularly for determining the [H3O+] and pH of the acid 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+]. To calculate pH, one must first determine the [H3O+] from the dissociation of the acid, which may involve using the Ka values and the initial concentration of the acid. Understanding this relationship is key to solving the problem presented.