Answer:
The final pH is 3.80
Explanation:
Step 1: Data given
Volume of acetic acid = 200.0 mL = 0.200 L
Number of moles acetic acid = 0.5000 moles
Volume of NaOH = 100.0 mL = 0.100 L
Molarity of NaOH = 0.500 M
Ka of acetic acid = 1.770 * 10^-5
Step 2: The balanced equation
CH3COOH + NaOH → CH3COONa + H2O
Step 3: Calculate moles
moles = molarity * volume
Moles NaOH = 0.500 M * 0.100 L
Moles NaOH = 0.0500 moles
Step 4: Calculate the limiting reactant
For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O
NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles
There will be produced 0.0500 moles CH3COONa
Step 5: Calculate the total volume
Total volume = 200.0 mL + 100.0 mL = 300.0 mL
Total volume = 0.300 L
Step 6: Calculate molarity
Molarity = moles / volume
[CH3COOH] = 0.450 moles / 0.300 L
[CH3COOH] = 1.5 M
[CH3COONa] = 0.0500 moles / 0.300 L
[CH3COONa]= 0.167 M
Step 7: Calculate pH
pH = pKa + log[A-]/ [HA]
pH = -log(1.77*10^-5) + log (0.167/ 1.5)
pH = 4.75 + log (0.167/1.5)
pH = 3.80
The final pH is 3.80
