A volume of 25.0 mL of 0.100 M CH3CO2H is titrated with 0.100 M NaOH. What is the pH after the addition of 12.5 mL of NaOH? (Ka for CH3CO2H = 1.8 x 10-5)
3.74
To find the pH after the addition of 12.5 mL of NaOH, we need to determine the number of moles of CH3CO2H and NaOH involved in the reaction.
Step 1: Calculate the moles of CH3CO2H initially present.
Molarity (M) is defined as moles of solute per liter of solution. Given the volume of CH3CO2H as 25.0 mL and the molarity as 0.100 M, we can calculate the moles of CH3CO2H:
moles of CH3CO2H = volume (in liters) × molarity
moles of CH3CO2H = 0.025 L × 0.100 M = 0.0025 moles
Step 2: Calculate the moles of NaOH added.
Given the volume of NaOH as 12.5 mL and the molarity as 0.100 M, we can calculate the moles of NaOH:
moles of NaOH = volume (in liters) × molarity
moles of NaOH = 0.0125 L × 0.100 M = 0.00125 moles
Step 3: Determine which species is limiting.
To determine the limiting reactant, we compare the moles of CH3CO2H and NaOH. As per the balanced equation, the ratio of CH3CO2H to NaOH is 1:1.
Since the moles of NaOH (0.00125 moles) are less than the moles of CH3CO2H (0.0025 moles), NaOH is the limiting reactant.
Step 4: Determine the moles of CH3CO2H and NaOH remaining after the reaction.
Since NaOH is the limiting reactant, it will react completely with the CH3CO2H. Thus, the moles of CH3CO2H remaining after the reaction will be:
moles of CH3CO2H remaining = moles of CH3CO2H initially present - moles of NaOH added
moles of CH3CO2H remaining = 0.0025 moles - 0.00125 moles = 0.00125 moles
Step 5: Calculate the concentration of CH3CO2H and CH3CO2^- after neutralization.
The volume of the final solution, after the addition of 12.5 mL of NaOH, will be the sum of the initial volumes of CH3CO2H and NaOH: 25.0 mL + 12.5 mL = 37.5 mL.
Converting the final volume to liters:
volume of final solution = 37.5 mL = 0.0375 L
The concentration of CH3CO2H (undissociated acid) after neutralization is given by:
concentration of CH3CO2H = moles of CH3CO2H remaining / volume of final solution
concentration of CH3CO2H = 0.00125 moles / 0.0375 L ≈ 0.033 M
The concentration of CH3CO2^- (conjugate base of CH3CO2H) after neutralization is equal to the concentration of NaOH added:
concentration of CH3CO2^- = moles of NaOH added / volume of final solution
concentration of CH3CO2^- = 0.00125 moles / 0.0375 L ≈ 0.033 M
Step 6: Calculate the pOH and pH of the solution.
Since CH3CO2H is a weak acid, we assume it undergoes partial ionization in water. The balanced equation for the ionization of CH3CO2H is:
CH3CO2H + H2O ⇌ CH3CO2^- + H3O+
From the equation, we see that the concentration of H3O+ (the hydronium ion) is equal to the concentration of CH3CO2^-.
To calculate the pOH of the solution, we use the concentration of CH3CO2^-:
pOH = -log10(concentration of CH3CO2^-)
pOH = -log10(0.033) ≈ 1.48
To find the pH, we use the equation:
pH = 14 - pOH
pH = 14 - 1.48 ≈ 12.52
Therefore, the pH after the addition of 12.5 mL of NaOH is approximately 12.52.