What is the pH of a solution that consists of 0.50M H2C6H6O6(ascorbic acid) and 0.75M NaHC6H6O6(sodium ascorbate)? for ascorbic acid, Ka=6.8x10-5
To find the pH of the solution, we need to determine the concentration of H+ ions in the solution.
In this case, we have a solution containing both ascorbic acid (H2C6H6O6) and sodium ascorbate (NaHC6H6O6). Ascorbic acid is a weak acid and can donate a proton (H+) to form the ascorbate ion (C6H6O6-) in solution. Sodium ascorbate can accept a proton (H+) to form sodium ion (Na+) and ascorbate ion (C6H6O6-).
The ascorbic acid will partially dissociate in water according to the following equation:
H2C6H6O6 ⇌ H+ + HC6H6O6-
The equilibrium constant (Ka) can be used to calculate the concentration of H+ ions.
Ka = [H+][HC6H6O6-]/[H2C6H6O6]
Since we know the value of Ka (6.8x10^-5) and the initial concentration of ascorbic acid (0.50M), we can set up an equilibrium expression:
6.8x10^-5 = [H+][0.50]/[H2C6H6O6]
However, we also need to consider the reaction between sodium ascorbate and water, where it accepts a proton to form ascorbate ion and hydroxide ion (OH-):
NaHC6H6O6 + H2O ⇌ Na+ + HC6H6O6- + OH-
The concentration of OH- ions will affect the pH of the solution. But as sodium hydroxide (NaOH) is a strong base, it will completely dissociate into Na+ and OH-. Since the concentration of sodium hydroxide is not given, we can assume it is negligible and ignore its contribution to the pH.
Therefore, we can consider the concentration of H+ ions only from the dissociation of ascorbic acid.
To calculate the concentration of H+ ions ([H+]), we need to solve the equilibrium expression for [H+].
Using rearrangement and simplification, we have:
[H+] = √(Ka * [H2C6H6O6])
Substituting the values, we get:
[H+] = √(6.8x10^-5 * 0.50)
[H+] ≈ 0.0041M
Since pH is defined as the negative logarithm of the concentration of H+ ions, we can calculate the pH of the solution using the formula:
pH = -log[H+]
Substituting the value, we get:
pH = -log(0.0041)
pH ≈ 2.39
Therefore, the pH of the solution is approximately 2.39.
To find the pH of the solution, we need to determine the concentration of H⁺ ions in the solution.
First, let's write the dissociation equation for ascorbic acid (H2C6H6O6):
H2C6H6O6 ⇌ H⁺ + HC6H6O6⁻
We can assume that ascorbic acid (H2C6H6O6) will partially dissociate, forming H⁺ and HC6H6O6⁻ ions.
Now, let's calculate the concentration of H⁺ ions from the dissociation of ascorbic acid.
Given:
- Concentration of ascorbic acid (H2C6H6O6) = 0.50 M
- Ka (acid dissociation constant) for H2C6H6O6 = 6.8x10⁻⁵
Let's assume that x is the concentration of H⁺ ions formed from the dissociation of H2C6H6O6.
[H⁺] = x M
[HC6H6O6⁻] = x M
Using the equilibrium expression for the dissociation of ascorbic acid:
Ka = [H⁺] [HC6H6O6⁻] / [H2C6H6O6]
Substituting the given values:
6.8x10⁻⁵ = x * x / (0.50 - x)
Since the value of x is expected to be small compared to 0.50 (as only partial dissociation occurs), we can neglect it in the denominator:
6.8x10⁻⁵ ≈ x * x / 0.50
Rearranging the equation:
x² = 6.8x10⁻⁵ * 0.50
x² = 3.4x10⁻⁵
x ≈ √(3.4x10⁻⁵)
x ≈ 0.00583 M
The concentration of H⁺ ions (x) is approximately 0.00583 M.
Now, let's calculate the pH of the solution using the concentration of H⁺ ions:
pH = -log[H⁺]
pH = -log(0.00583)
pH ≈ -(-2.2347)
pH ≈ 2.235
Therefore, the pH of the solution consisting of 0.50 M H2C6H6O6 (ascorbic acid) and 0.75 M NaHC6H6O6 (sodium ascorbate) is approximately 2.235.
This is almost like your other buffer posts. Use the Henderson-Hasselbalch equation.
pH = pKa + logp[(base)/(acid)]
The base is the ascorbate salt and the acid is ascorbic acid.
Post your work if you get stuck.