What is the pH of a 0.001 M solution of acetylsalicylic acid (aspirin) (Ka 3.0 x 10-4)?
Denote the chemical formula for the acid to be HA, then
Ka=[H+][A-]/[HA]=3.0*10-4
at a given temperature.
ACE table for
H+ A- HA
A 0 0 0.001
C x x -x
E x x 0.001-x
At equilibrium,
3.0*10-4=x²/(0.001-x)
Cross multiply to get
3.0*10-4(0.001-x)=x²
Solve for x, reject negative root.
and pH is -log10(x).
It should be between 3 and 3.5
(typo in format)
Cross multiply to get
3.0*10-4(0.001-x)=x²
Solve for x, reject negative root.
and pH is -log10(x).
It should be between 3 and 3.5.
Can't seem to get it right!
the -4 should be in superscript.
To find the pH of a solution of acetylsalicylic acid (aspirin), we need to consider the acid dissociation of the aspirin molecule. Acetylsalicylic acid is a weak acid, which means it will only partially dissociate in water.
The first step is to write down the balanced chemical equation for the dissociation of acetylsalicylic acid:
C9H8O4(aq) + H2O(l) ⇌ C9H7O4-(aq) + H3O+(aq)
In this reaction, acetylsalicylic acid (C9H8O4) donates a proton (H+) to water, forming the acetate ion (C9H7O4-) and the hydronium ion (H3O+).
The next step is to find the expression for the acid dissociation constant (Ka). Ka is the ratio of the concentration of products to reactants at equilibrium.
Ka = [C9H7O4-][H3O+] / [C9H8O4]
Given the Ka value for acetylsalicylic acid (3.0 x 10^-4) and the fact that the initial concentration of acetylsalicylic acid is 0.001 M, we can assume that the concentration of the hydronium ion (H3O+) at equilibrium is equal to the concentration of the ionized acetylsalicylic acid ([C9H7O4-]).
Using the quadratic equation or approximation methods, we can solve for [H3O+] and determine the pH of the solution.
However, it is worth noting that the concentration of acetylsalicylic acid is relatively high compared to its Ka value, indicating that the dissociation will be relatively small. Therefore, we can approximate the concentration of the hydronium ion as the square root of Ka multiplied by the initial concentration of acetylsalicylic acid.
[H3O+] = √(Ka x [C9H8O4])
[H3O+] = √(3.0 x 10^-4 x 0.001)
[H3O+] = √(3.0 x 10^-7)
[H3O+] = 5.48 x 10^-4 M
Now, we can find the pH of the solution by taking the negative logarithm (base 10) of the hydronium ion concentration:
pH = -log[H3O+]
pH = -log(5.48 x 10^-4)
pH ≈ 3.26
Therefore, the pH of a 0.001 M solution of acetylsalicylic acid (aspirin) is approximately 3.26.