The amino acid glycine (H 2 N–CH 2 –COOH) can participate in the following equilibria in water: H 2 N–CH 2 –COOH + H 2 O ⇌ H 2 N–CH 2 –COO – + H 3 O + K a = 4.3 × 10 -3 H 2 N–CH 2 –COOH + H 2 O⇌ + H 3 N–CH 2 –COOH + OH - K b = 6.0 × 10 -5 (b) What is the pH of a 0.050 M aqueous solution of glycine?

Hi everyone. This problem reads the equilibrium of the amino acid align and water are the following, calculate the P. H. Of a 0.9 moller alkaline solution. Okay, so we want to calculate ph and we know we're dealing with an acid. So we need to first determine whether or not this is a strong acid or a weak acid because that's going to determine how we solve this problem. So this is a weak acid. Okay. Which means when it dissociates it's not going to dissociate completely into its ions. Okay. And so with ph because this is an acid with an acid, an acid is going to produce hydro knee, um ions and P H is equal to the negative log of the concentration of those hydro ni um ions. Ok, so essentially we're going to need to figure out what is our concentration of hydrogen ions. And in the problem, something else they give us are the K and K B, which are the acid dissociation constant and based association constant. So for R K A R K A is going to be equal to K W over K B. We're going to need to solve for it because the value that's given is not the final K A. So our KW is a constant. We should know which is 1.0 times to the negative 14. And the KB which was what was given is what we'll use. So we have 7.41 times 10 to the negative five. So once we calculate this we get our K A. Is equal to 1. times 10 to the negative 10. We're going to need to know this to solve for a concentration of hydrogen ions. Okay, so let's go ahead and write out our K. A. Expression. R K expression is equal to the concentration of our products over our reactant. Okay. So looking at our equation or reaction, we see we're going to take copy down what the product of our the product of our products. Okay. So we have two products. Mhm. Yeah. Mhm. Okay. And our reactant. So for our reactant. So we're looking at this one here because this is where we have our hydro knee um ions. So for R K. K. Expression remember we don't include solids or liquids and our water here is a liquid. So we're not going to include it. So we're only going to include our valentin which is our reactant and it's going to be by itself. Okay, so our concentration of products over our concentration of reactant. And remember that we're trying to solve for our concentration of hydro ni um ions here so that we can plug it in to here so we can solve for P. H. All right. So what's next is we're going to substitute in X. Okay. For our products So X times X. And these concentrations are going to be equal to each other and for our reactant it's going to be our initial concentration which they tell us is 0.9 and we're going to do minus X. Okay because that's going to tell us what are reactant concentration is and we would see this if we did an ice table. Alright so simplifying this or writing it out we're going to get our we're gonna have X times X over zero 0.9 minus X. Is equal to our K. A. Which is 1.3495 times 10 To the -10. Okay so what we can do now is we have a minus X. In the denominator. We can check to see if this X. Is negligible and if it is negligible we can ignore it but we need to check first to see is it negligible. Okay so let's write that down is X negligible. And the way we do that is by taking our initial concentration which is 0.9 and dividing it by R. K. A. Value which we saw four. So 1.3495 times 10 to the negative 10. If the number you get is greater than 500 then X. Is negligible. If it's less than 500 then X is not negligible. So here it is greater than 500. So yes X is negligible. So we're going to go ahead and ignore it. So let's just simplify what we're left with here. So we're left with X squared over 0.9 is equal to 1.3495 times 10 to the negative 10. Now we can solve for X. So let's go ahead and simplify here. So we have X squared is equal to 1.3495 times 10 to the negative 10 times 0.9. Okay so we get X squared is equal to 1.21455 times to the negative 11 to get rid of the squared on the left side. We're going to take the square root of both sides. So we get X. Is equal to the square root of one point 21455 times 10 to the negative 11. Which leaves us with X is equal to 3. times 10 to the negative six. And recall X is equal to our concentration of hydro ni um ions. And we said we know that P. H. Is equal to the negative log of our hydro knee um concentration. So we have what we need to solve for P. H. We just need to plug in our value for hydro knee um concentration which we just solved for. Okay, so we get a final answer of P. H. is equal to 5. and that is our final answer. That's it for this problem. I hope this was helpful

Ch.16 - Acid-Base Equilibria

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Brown 14th Edition

Ch.16 - Acid-Base Equilibria

Problem 114b

Chapter 16, Problem 114b

The amino acid glycine (H₂N–CH₂–COOH) can participate in the following equilibria in water:
H₂N–CH₂–COOH + H₂O ⇌ H₂N–CH₂–COO^– + H₃O⁺ K_a = 4.3 × 10^-3
H₂N–CH₂–COOH + H₂O⇌ ⁺H₃N–CH₂–COOH + OH^- K_b = 6.0 × 10^-5
(b) What is the pH of a 0.050 M aqueous solution of glycine?

Verified step by step guidance

Identify the relevant equilibria for glycine in water: the acid dissociation (Ka) and the base dissociation (Kb).

Recognize that glycine can act as both an acid and a base, making it an amphoteric species. Use the given Ka and Kb values to determine which equilibrium will dominate in a 0.050 M solution.

Calculate the concentration of hydronium ions \([H_3O^+]\) using the acid dissociation constant \(K_a\) and the initial concentration of glycine. Use the expression: \[ K_a = \frac{[H_3O^+][H_2N-CH_2-COO^-]}{[H_2N-CH_2-COOH]} \]

Assume that the change in concentration of glycine due to dissociation is small compared to the initial concentration, allowing simplification of the equilibrium expression.

Calculate the pH from the hydronium ion concentration using the formula: \( \text{pH} = -\log[H_3O^+] \).

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Key Concepts

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

Acid-Base Equilibria

Acid-base equilibria involve the transfer of protons (H+) between species in solution. In the case of glycine, it can act as both an acid and a base, participating in equilibria that define its behavior in water. Understanding the dissociation constants (Ka and Kb) is crucial for predicting the pH of the solution, as they indicate the strength of the acid and base forms of glycine.

Dissociation Constants (Ka and Kb)

Dissociation constants, Ka and Kb, quantify the extent to which an acid or base dissociates in water. Ka represents the equilibrium constant for the dissociation of an acid, while Kb represents the equilibrium constant for the dissociation of a base. For glycine, the given values of Ka and Kb allow us to calculate the concentrations of the various species in solution, which is essential for determining the pH.

pH Calculation

pH is a measure of the hydrogen ion concentration in a solution, calculated using the formula pH = -log[H+]. In the context of glycine, the pH can be determined by considering the contributions of both the acidic and basic forms of the amino acid in solution. By applying the Henderson-Hasselbalch equation or directly using the dissociation constants, one can find the pH of the glycine solution.

The amino acid glycine (H2N–CH2–COOH) can participate in the following equilibria in water: H2N–CH2–COOH + H2O ⇌ H2N–CH2–COO– + H3O+ Ka = 4.3 × 10-3 H2N–CH2–COOH + H2O⇌ +H3N–CH2–COOH + OH- Kb = 6.0 × 10-5 (b) What is the pH of a 0.050 M aqueous solution of glycine?

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Key Concepts

Acid-Base Equilibria

Dissociation Constants (Ka and Kb)

pH Calculation