a buffer prepared by mixing equal moles of an acid having Ka=4.5x10^-4 and a salt of its conjugate base has a pH of what?
One thing to learn quickly in buffers is that when bae = acid(as in this case) is that pH = pKa. You know Ka so -log Ka = pKa = pH
To determine the pH of a buffer prepared by mixing equal moles of an acid and its conjugate base salt, we need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of the acid and the ratio of the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([conjugate base] / [acid])
In this case, the Ka (acid dissociation constant) of the acid is given as 4.5x10^-4.
Since equal moles of the acid and its conjugate base salt are mixed, the ratio of their concentrations will be 1.
Now, let's calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log ([conjugate base] / [acid])
= log (1)
= 0
Therefore, the pH of the buffer prepared by mixing equal moles of the acid and its conjugate base salt is 0.
To determine the pH of a buffer solution prepared by mixing equal moles of an acid and its conjugate base, you need to consider the equilibrium between the acid and its conjugate base.
In this case, the acid has a Ka value of 4.5x10^-4. Ka is the acid dissociation constant and is a measure of the extent to which an acid dissociates in water. The larger the Ka value, the stronger the acid.
Since you have equal moles of the acid and its conjugate base, their concentrations will be the same in the buffer solution. Let's assume their concentration is denoted by [A] (for the acid) and [B] (for the conjugate base) in moles per liter (M).
The equilibrium expression for the dissociation of the acid can be written as:
Ka = [H+][A-] / [HA]
Since you have equal moles of the acid and its conjugate base, the initial concentration of the acid ([HA]) will be equal to the initial concentration of the conjugate base ([A-]).
Therefore, the equilibrium expression simplifies to:
Ka = [H+]^2 / [A]
Rearranging this equation, we get:
[H+]^2 = Ka * [A]
Taking the square root of both sides, we obtain:
[H+] = √(Ka * [A])
To calculate the pH, which is a measure of the concentration of hydrogen ions ([H+]), we will use the equation:
pH = -log[H+]
Now, let's plug in the values. In this case, we have equal moles, so let's assume the initial concentration of the acid and its conjugate base is 1 M. Therefore, [A] = [B] = 1 M.
[H+] = √(Ka * [A])
[H+] = √(4.5x10^-4 * 1)
[H+] ≈ 6.71x10^-3 M
Calculating the pH:
pH = -log[H+]
pH = -log(6.71x10^-3)
pH ≈ 2.17
Therefore, the pH of the buffer solution prepared by mixing equal moles of the acid and its conjugate base is approximately 2.17.