Calculate the approximate pH of a 0.1 M solution of ethanoic acid at 298 K, given that the
dissociation constant, Ka, is 1.75 x 10-5 mol/L at 298 K.Ethanoic acid dissociates in aqueous solution as follows:
CH3COOH(aq) + H2O H3O+(aq) + CH3COO-(aq)
You need to find the arrow key on your keyboard and use it. I will use acetic acid (ethanoic acid) as HAc.
.........HAc + H2O ==> H3O^+ + Ac^-
I........0.1...........0........0
C.........-x...........x........x
E.......0.1-x..........x........x
Substitute the E line into Ka expression and solve for x = (H3O^+) and convert that to pH.
To calculate the pH of a solution of ethanoic acid, we need to use the dissociation constant (Ka) and the concentration of the acid.
Step 1: Write the balanced equation for the dissociation of ethanoic acid:
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq)
Step 2: Set up the equilibrium expression for Ka:
Ka = [H3O+][CH3COO-] / [CH3COOH]
Step 3: Substitute the given value of Ka (1.75 × 10^(-5) mol/L), and let's assume that the concentration of [H3O+] is x.
1.75 × 10^(-5) = x * x / (0.1 - x)
Step 4: Solve the equation for x.
Using the quadratic formula, we can solve for x:
x = [-b ± √(b^2 - 4ac)] / (2a)
For our equation:
a = 1
b = 0
c = -1.75 × 10^(-5)
x = [-0 ± √(0 - 4 * 1 * (-1.75 × 10^(-5)))] / (2 * 1)
x = ± √(7 × 10^(-5)) / 2
Since we are dealing with a concentration, the negative solution does not make sense, so we take the positive solution:
x = √(7 × 10^(-5)) / 2
Step 5: Calculate the pH using the equation pH = -log[H3O+].
pH = -log(√(7 × 10^(-5)) / 2)
Using the logarithmic identities, we can simplify the equation:
pH = -log(√(7 × 10^(-5))) - log(2)
pH = -0.5 * log(7 × 10^(-5)) - log(2)
pH ≈ -0.5 * (-4.154) - 0.301
pH ≈ 2.077 + 0.301
pH ≈ 2.378
Therefore, the approximate pH of a 0.1 M solution of ethanoic acid at 298 K is 2.378.
To calculate the approximate pH of the ethanoic acid solution, we need to determine the concentration of the hydronium ion (H3O+) in the solution.
Step 1: Write the dissociation equation of ethanoic acid:
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO-(aq)
The stoichiometry of the reaction indicates that for every molecule of ethanoic acid that dissociates, one hydronium ion (H3O+) is formed. Therefore, the concentration of the hydronium ion is equal to the concentration of the dissociated ethanoic acid.
Step 2: Calculate the concentration of hydronium ion:
Given that the concentration of ethanoic acid is 0.1 M, the concentration of hydronium ion is also 0.1 M.
Step 3: Calculate the pH:
The pH of a solution can be calculated using the formula:
pH = -log[H3O+]
Using this formula, we can find the pH of the solution:
pH = -log(0.1) = -(-1) = 1
Therefore, the approximate pH of the 0.1 M solution of ethanoic acid is 1.