What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M CH3CO2H with 25.00 mL of 0.030 M CH3CO2Na? Assume that the volume of the solutions are additive and that a = 1.8 × 10-5 for CH3CO2H.
To determine the pH of the solution, we need to calculate the concentration of the hydronium ion (H3O+). We can do this by considering the acid-base reaction between CH3CO2H and CH3CO2Na.
The balanced chemical equation for the reaction is:
CH3CO2H + H2O ⇌ CH3CO2- + H3O+
From the given information, we know the initial concentrations of CH3CO2H and CH3CO2Na, and we can assume that they completely dissociate. Therefore, we have:
[CH3CO2H] = 0.10 M
[CH3CO2Na] = 0.030 M
Since the volume of the solutions is additive, the total volume of the resulting solution is 50.00 mL.
Now, let's set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations at equilibrium. Assuming x is the concentration of H3O+:
CH3CO2H + H2O ⇌ CH3CO2- + H3O+
Initial: 0.10 M - - - x M
Change: -x - - +x +x M
Equilibrium: 0.10 - x - - x x M
According to the ICE table, the equilibrium concentration of CH3CO2H is 0.10 - x M, and the equilibrium concentration of H3O+ is x M.
We also know that the acid dissociation constant for CH3CO2H (Ka) is 1.8 × 10^-5.
Using the equilibrium concentrations in the expression for Ka, we can write:
Ka = [CH3CO2-][H3O+] / [CH3CO2H]
Substituting the equilibrium concentrations:
1.8 × 10^-5 = (x)(x) / (0.10 - x)
At this point, we can simplify the equation by assuming x is much smaller than 0.10 (a valid assumption for weak acids). Therefore, we can approximate 0.10 - x as 0.10:
1.8 × 10^-5 = x^2 / 0.10
Rearranging the equation:
x^2 = (1.8 × 10^-5)(0.10)
x = √(1.8 × 10^-5)(0.10)
x ≈ 0.0042 M
Now that we have the concentration of H3O+, we can calculate the pH.
pH = -log[H3O+]
pH = -log(0.0042)
pH ≈ 2.38
Therefore, the pH of the solution prepared by mixing 25.00 mL of 0.10 M CH3CO2H with 25.00 mL of 0.030 M CH3CO2Na is approximately 2.38.
To determine the pH of the solution, we need to calculate the concentration of the hydrogen ion (H+) in the final solution. Here's how you can do it step-by-step:
Step 1: Determine the moles of CH3CO2H and CH3CO2Na in the solution.
- Moles of CH3CO2H = volume (L) × concentration (M) = (0.025 L) × (0.10 M) = 0.0025 moles
- Moles of CH3CO2Na = volume (L) × concentration (M) = (0.025 L) × (0.030 M) = 0.00075 moles
Step 2: Calculate the moles of the acid and base that remain dissociated.
- Moles of CH3CO2H remaining = moles of CH3CO2H - moles of CH3CO2Na = 0.0025 - 0.00075 = 0.00175 moles
Step 3: Calculate the moles of H+ ions formed from the dissociation of CH3CO2H.
- Moles of H+ formed = moles of CH3CO2H remaining × α (dissociation constant) = 0.00175 moles × 1.8 × 10^-5 = 3.15 × 10^-8 moles
Step 4: Calculate the total volume of the solution.
- Total volume = volume of CH3CO2H + volume of CH3CO2Na = 0.025 L + 0.025 L = 0.05 L
Step 5: Calculate the concentration of H+ ions.
- Concentration of H+ ions = moles of H+ formed / total volume = 3.15 × 10^-8 moles / 0.05 L = 6.3 × 10^-7 M
Step 6: Calculate the pH of the solution using the formula pH = -log[H+].
- pH = -log(6.3 × 10^-7) ≈ 6.20