Ascorbic C3H7O4COOH,also known as VitaminC, is an essential Vitamin for all mammals. Among mammals, only humans, monkeys and guinea pigs cannot synthesis it in their bodies LA for ascorbic acid is 7.9*10^-5. Calculate [H3O+] and pH in a 0.110M solution of the acid.
Did you mean synthesize? Did you mean Ka = 7.9E-5?
Call ascorbic acid Has, then
...............Has ==> H^+ + as^-
I.............0.110........0..........0
C............-x..............x..........x
E.........0.110-x........x...........x
Write the Ka expression for ascorbic acid, substitute the E line above and solve for H^+ = (H3O^+). Convert that to pH by pH = -log(H^+).
Post your work if you get stuck.
To calculate the [H3O+] and pH of a 0.110M solution of ascorbic acid (C3H7O4COOH), we can use the expression for the acid dissociation constant (Ka) and the definition of pH.
First, let's write the balanced equation for the dissociation of ascorbic acid:
C3H7O4COOH + H2O ⇌ C3H7O4COO- + H3O+
The acid dissociation constant (Ka) for ascorbic acid is given as 7.9*10^-5. This can be used to express the equilibrium constant for the dissociation reaction as:
Ka = [C3H7O4COO-][H3O+] / [C3H7O4COOH]
Since the initial concentration of ascorbic acid (C3H7O4COOH) is 0.110M, we can assume that the concentration of the dissociated form (C3H7O4COO-) is negligible compared to the initial concentration.
Therefore, we can simplify the equation to:
Ka = [H3O+] / [C3H7O4COOH]
Plugging in the values, we have:
7.9*10^-5 = [H3O+] / 0.110
Rearranging the equation to solve for [H3O+], we get:
[H3O+] = 7.9*10^-5 * 0.110 = 8.69*10^-6
Now, let's calculate the pH using the definition of pH as:
pH = -log[H3O+]
Plugging in the value for [H3O+], we have:
pH = -log(8.69*10^-6) = 5.06
Therefore, the [H3O+] is approximately 8.69*10^-6 and the pH is approximately 5.06 in a 0.110M solution of ascorbic acid.
To calculate the [H3O+] and pH in a solution of ascorbic acid, we first need to understand that ascorbic acid is a weak acid, meaning that it partially dissociates in water.
First, let's write the balanced chemical equation for the dissociation of ascorbic acid:
C3H7O4COOH ⇌ C3H7O4COO- + H+
Given that the equilibrium constant (Ka) for ascorbic acid is 7.9 * 10^-5, we can determine the concentration of [H3O+] by using the equation:
Ka = [C3H7O4COO-][H+] / [C3H7O4COOH]
Since the initial concentration of ascorbic acid ([C3H7O4COOH]) is 0.110M, and assuming that x is the concentration of [C3H7O4COO-] and [H+], we can write the equilibrium expression as:
(7.9 * 10^-5) = (x)(x) / (0.110 - x)
Next, we can approximate that x is very small compared to 0.110, so we can ignore x in the denominator:
(7.9 * 10^-5) = (x)(x) / (0.110)
Simplifying further:
x^2 = (7.9 * 10^-5) * (0.110)
x^2 = 8.69 * 10^-6
Taking the square root of both sides:
x ≈ √(8.69 * 10^-6)
x ≈ 9.33 * 10^-3
Now that we have the concentration of [H+], we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(9.33 * 10^-3)
pH ≈ 2.03
Therefore, in a 0.110M solution of ascorbic acid, the [H3O+] is approximately 9.33 * 10^-3 M and the pH is approximately 2.03.