What is the pH of a solution obtained by dissolving 0.325 g of acetylsalicylic acid, in 0.5 L of water? With a Ka of 3.3 x 10 ^-4.
To find the pH of the solution, we need to determine the concentration of the hydronium ions (H3O+) in the solution.
First, we need to calculate the concentration of acetylsalicylic acid (ASA) in the solution.
Given:
Mass of ASA = 0.325 g
Volume of water = 0.5 L
We first convert the mass of ASA to moles. The molar mass of acetylsalicylic acid (ASA) is 180.16 g/mol.
Moles of ASA = mass / molar mass
= 0.325 g / 180.16 g/mol
= 0.001805 mol
Now, we calculate the concentration of ASA in the solution:
Concentration (M) = moles / volume
= 0.001805 mol / 0.5 L
= 0.00361 M
Since acetylsalicylic acid is a weak acid, it undergoes partial ionization in water. The equilibrium equation for the ionization is:
ASA (aq) + H2O (l) ⇌ H3O+ (aq) + A- (aq)
The equilibrium constant (Ka) is given as 3.3 x 10^-4.
To solve for the concentration of H3O+ (aq), we can assume that the concentration of ASA that ionizes is small compared to the initial concentration (0.00361 M). Therefore, we can neglect the change in ASA concentration due to ionization.
Using the Ka expression:
Ka = [H3O+][A-] / [ASA]
Let x be the concentration of H3O+ and A- ions. Since the concentration of H3O+ is equal to the concentration of A-, we can simplify it to:
Ka = x^2 / (0.00361 - x)
At equilibrium, the change in concentration of ASA is negligible, so we can assume x is small compared to 0.00361 M. This simplifies the equation to:
Ka = x^2 / 0.00361
Rearranging and solving for x:
x^2 = Ka * 0.00361
x = √(Ka * 0.00361)
Substituting the given Ka value:
x = √(3.3 x 10^-4 * 0.00361)
x ≈ 7.11 x 10^-4 M
Now, to find the pH, we'll use the formula:
pH = -log[H3O+]
pH = -log(7.11 x 10^-4)
pH ≈ 3.15
Therefore, the pH of the solution obtained by dissolving 0.325 g of acetylsalicylic acid in 0.5 L of water is approximately 3.15.
To calculate the pH of a solution obtained by dissolving acetylsalicylic acid in water, we need to determine the concentration of the resulting acetylsalicylic acid solution.
Step 1: Calculate the number of moles of acetylsalicylic acid:
Molar mass of acetylsalicylic acid (C9H8O4) = (9 * 12.01) + (8 * 1.01) + (4 * 16.00) = 180.16 g/mol
Number of moles of acetylsalicylic acid = mass / molar mass
Number of moles = 0.325 g / 180.16 g/mol
Step 2: Calculate the concentration of acetylsalicylic acid solution:
Concentration (in mol/L) = moles / volume (in L)
Concentration = (0.325 g / 180.16 g/mol) / 0.5 L
Step 3: Calculate the concentration of acetylsalicylic acid (HAc) that ionizes:
Assuming complete ionization of HAc, the concentration of H+ ions would be the same as the initial concentration of acetylsalicylic acid. So, [H+] = concentration of HAc
Step 4: Calculate the concentration of Acetyl Salicylate (Ac-) ions:
Since acetylsalicylic acid is monoprotic, the concentration of Ac- ions would be equal to the concentration of H+ ions.
Step 5: Write the equilibrium expression for the ionization of acetylsalicylic acid:
HAc ⇌ H+ + Ac-
The equilibrium constant expression (Ka) is given as:
Ka = [H+][Ac-] / [HAc]
Step 6: Substitute the given values into the equilibrium constant expression:
Ka = [H+][H+] / [HAc] = [H+]^2 / [HAc]
Step 7: Rearrange the equilibrium constant expression:
[H+]^2 = Ka * [HAc]
Step 8: Solve for [H+]:
[H+] = sqrt(Ka * [HAc])
Step 9: Calculate the pH:
pH = -log[H+]
Now, you can substitute the values into the equation and calculate the pH.