A 0.010 M solution of aspirin, a weak monoprotic acid, has a pH of 3.3. What is the Ka of aspirin?
To determine the Ka of aspirin, we need to use the pH value of the solution and the concentration of the acid.
First, let's understand the definition of Ka. Ka is the acid dissociation constant, which measures the extent to which an acid dissociates or ionizes in water.
The general expression for the dissociation of a weak monoprotic acid, like aspirin (HA), is represented as follows:
HA ⇌ H+ + A-
In this case, aspirin (HA) acts as the acid while H+ represents the hydronium ion and A- represents the conjugate base.
The Ka expression for this dissociation reaction is:
Ka = [H+][A-] / [HA]
Here, [H+] represents the concentration of hydronium ions, [A-] represents the concentration of the conjugate base, and [HA] represents the concentration of the acid.
Given that the concentration of the aspirin solution is 0.010 M and the pH is 3.3, we know that the concentration of hydronium ions ([H+]) is equal to 10^(-pH).
[H+] = 10^(-pH)
[H+] = 10^(-3.3)
Calculating 10^(-3.3) gives us approximately 5.01 x 10^(-4) M.
Based on the Ka expression, we can plug in the known values:
Ka = [H+][A-] / [HA]
Ka = (5.01 x 10^(-4))(A-) / 0.010
We can simplify by multiplying both sides by 0.010 to isolate (A-):
Ka * 0.010 = 5.01 x 10^(-4)(A-)
0.010 * Ka = 5.01 x 10^(-4)(A-)
Finally, we divide both sides by 5.01 x 10^(-4) to solve for Ka:
Ka = (0.010 * Ka) / (5.01 x 10^(-4))
By calculating this equation, we can find the value of Ka for aspirin.