0.100 M NaOH (in the burette) is titrated with 25.0 mL 0.120 M CH3COOH (pKa = 4.75) in a conical flask. Determine the pH of the solution in the flask after 10 mL of the NaOH has been added.
millimols CH3COOH = 25.0 x 0.12 = 3.0
millimols NaOH = 10 x 0.1 = 1
..CH3COOH + NaOH ==> CH3COONa + H2O
I...3.0......0........0..........0
add..........1.0.................
C...-1.0....-1.0......+1.0.......
E....2.0......0.......1.0
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH.
pH = pKa + log (base)/(acid)
pH = 4.75 + log (1/35)/(2/35)
To determine the pH of the solution in the flask after adding 10 mL of NaOH, we first need to calculate the moles of CH3COOH and NaOH.
1. Calculate the moles of CH3COOH:
Moles of CH3COOH = Volume (L) x Concentration (M)
Moles of CH3COOH = 0.025 L x 0.120 M
Moles of CH3COOH = 0.003 moles
2. Calculate the moles of NaOH:
Moles of NaOH = Volume (L) x Concentration (M)
Moles of NaOH = 0.010 L x 0.100 M
Moles of NaOH = 0.001 moles
Now, we need to determine the limiting reagent to identify the reaction that will occur. In this case, NaOH and CH3COOH react in a 1:1 ratio according to the stoichiometry of the balanced equation:
CH3COOH + NaOH -> CH3COONa + H2O
Since the moles of NaOH (0.001 mol) are less than the moles of CH3COOH (0.003 mol), NaOH is the limiting reagent.
Now, let's calculate the remaining moles of CH3COOH after the reaction:
Remaining moles of CH3COOH = Initial moles of CH3COOH - Moles of NaOH used in reaction
Remaining moles of CH3COOH = 0.003 mol - 0.001 mol
Remaining moles of CH3COOH = 0.002 mol
Now, let's calculate the concentration of CH3COOH after the reaction:
Concentration of CH3COOH = Remaining moles of CH3COOH / Volume (L)
Concentration of CH3COOH = 0.002 mol / 0.015 L (10 mL + 5 mL initially)
Concentration of CH3COOH = 0.133 M
Next, we need to calculate the concentration of the acetate ion (CH3COO-) formed from the reaction:
Concentration of CH3COO- = Moles of CH3COOH reacted / Volume (L)
Concentration of CH3COO- = 0.001 mol / 0.015 L (10 mL + 5 mL initially)
Concentration of CH3COO- = 0.067 M
Finally, we can determine the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([CH3COO-] / [CH3COOH])
pH = 4.75 + log(0.067 M / 0.133 M)
pH = 4.75 - log 2
pH ≈ 4.75 - 0.301
pH ≈ 4.449
Therefore, the pH of the solution in the flask after adding 10 mL of NaOH is approximately 4.449.