What is the pH of a 5.00E-2 M solution of acetylsalicylic acid, pKa = 3.52.
...........ASA ==> H^+ + A^-
I..........0.05.....0.....0
C..........-x.......x......x
E.........0.05-x....x......x
K = 3.02E-4 = (H^+)(A^-)/(ASA)
Substitute from the ICE table into Ka and solve for (H^+), then convert to pH.
To determine the pH of a solution of acetylsalicylic acid, we need to consider its dissociation in water. Acetylsalicylic acid (ASA) is a weak acid that partially dissociates in water, releasing H+ ions.
The dissociation reaction of acetylsalicylic acid can be represented as follows:
ASA(aq) ⇌ H+(aq) + A-(aq)
The equilibrium constant (Ka) for this reaction is related to the acid dissociation constant (pKa) by the equation:
Ka = 10^(-pKa)
Now, let's calculate the value of Ka:
pKa = 3.52
Ka = 10^(-3.52)
= 2.51 x 10^(-4)
Next, we need to determine the concentration of H+ ions in the solution. Since acetylsalicylic acid is a weak acid, the concentration of H+ ions can be approximated as the same as the concentration of the dissociated acetylsalicylic acid.
So, the concentration of H+ ions is 5.00E-2 M.
Now, we can calculate the pH using the equation:
pH = -log[H+]
pH = -log(5.00E-2)
= -log(5.00) + log(10^(-2))
= -log(5.00) -2
= 1.30 - 2
= -0.70
Therefore, the pH of the 5.00E-2 M solution of acetylsalicylic acid is approximately -0.70.
To find the pH of a solution of acetylsalicylic acid, we need to consider the dissociation of the acid and its relationship to the pKa value.
The pKa value represents the acidity of a compound and is defined as the negative logarithm (base 10) of the acid dissociation constant (Ka). The lower the pKa value, the stronger the acid.
In this case, acetylsalicylic acid is a weak acid, and its dissociation can be represented by the following equation:
Acetylsalicylic acid (HA) ⇌ Acetylsalicylate ion (A-) + H+
The equilibrium constant (Ka) for this reaction is directly related to the acid and conjugate base concentrations. The dissociation constant (Ka) can be calculated from the pKa value using the equation:
Ka = 10^(-pKa)
In this case, the pKa value is given as 3.52, so we can calculate Ka as follows:
Ka = 10^(-3.52) = 3.0 x 10^(-4)
Now that we have the value of Ka, we can use it to calculate the concentration of H+ ions in the solution.
Let's assume the concentration of H+ ions in the solution at equilibrium is x mol/L. Then, the concentration of acetylsalicylic acid (HA) is (5.00 × 10^(-2) - x) mol/L, and the concentration of acetylsalicylate ion (A-) is x mol/L.
Using the equilibrium expression for Ka:
Ka = [A-][H+] / [HA]
Substituting the respective concentrations:
3.0 × 10^(-4) = x * x / (5.00 × 10^(-2) - x)
Simplifying the equation and rearranging, we get a quadratic equation:
x^2 - (3.0 × 10^(-4))(5.00 × 10^(-2) - x) = 0
Solving this equation will give us the concentration of H+ ions at equilibrium.
Once you have obtained this concentration, you can convert it into pH using the equation:
pH = -log [H+]
Plug in the value of [H+] to find the pH of the solution.