50 ml of 0.02M acetic acid is titrated with 0.1M NaOH.Calculate the pH of the solution when 10ml of NaOH is added
50 mL x 0.02M = 1 mmol HAc.
10 mL x 0.1M = 1 mmol NaOH.
HAc + NaOH ==> H2O + NaAc
1 mmol 1mmol = e.p.
So you have a solution of Ac^- in aqueous solution. The (Ac^-) is 1 mmol/60 mL = about 0.017M but you nee to do it more accurately.
........Ac^- + HOH ==> HAc + OH^-
I........0.017...........0.....0
C........-x.............x......x
E......0.017-x..........x......x
Kb for Ac^- = (Kw/Ka for HAc) = (HAc)(OH^-)/(Ac^-) and solve for x = OH^-, then convert to pH. I expect you will need to use the quadratic.
0.0012
To calculate the pH of the solution after adding 10 mL of NaOH, we need to determine the amount of acetic acid and its conjugate base (acetate ion) present in the solution at that point. We can then use the Henderson-Hasselbalch equation to find the pH.
Step 1: Calculate moles of acetic acid initially present.
Molarity (M) = moles/Liter
Moles (n) = Molarity (M) × Volume (L)
Moles of acetic acid = 0.02 M × 0.050 L = 0.001 moles
Step 2: Calculate moles of NaOH added.
Moles of NaOH = Molarity (M) × Volume (L)
Moles of NaOH = 0.1 M × 0.010 L = 0.001 moles
Step 3: Determine which reactant is limiting.
In a 1:1 acid-base reaction, the limiting reactant is the one with fewer moles. In this case, both acetic acid and NaOH have the same number of moles (0.001 moles).
So, neither reactant is in excess or deficient, and they will react until they are completely consumed. The reaction between acetic acid (CH3COOH) and sodium hydroxide (NaOH) can be represented as follows:
CH3COOH + NaOH -> CH3COONa + H2O
The reaction will produce acetate ion (CH3COO-) and water.
Step 4: Calculate the moles of NaOH reacted.
To find the moles of NaOH reacted, we consider the stoichiometry of the reaction, which shows a 1:1 ratio between acetic acid and NaOH.
Since both acetic acid and NaOH have the same number of moles (0.001 moles), all 0.001 moles of NaOH will react with 0.001 moles of acetic acid.
Step 5: Calculate the remaining moles of acetic acid and acetate ion.
Moles of acetic acid remaining = Initial moles of acetic acid (0.001 moles) - Moles of acetic acid reacted (0.001 moles) = 0 moles
Moles of acetate ion produced = Moles of acetic acid reacted (0.001 moles) = 0.001 moles
Step 6: Calculate the concentration of acetic acid and acetate ion.
Concentration (M) = moles/Volume (L)
Concentration of acetic acid = 0 moles / 0.050 L = 0 M (completely reacted)
Concentration of acetate ion = 0.001 moles / 0.050 L = 0.02 M
Step 7: Use the Henderson-Hasselbalch equation to calculate the pH.
pH = pKa + log([A-]/[HA])
The pKa of acetic acid is 4.76, which is the negative logarithm of its acid dissociation constant.
pH = 4.76 + log(0.02/0) = 4.76 (Since the concentration of acetic acid becomes 0)
Therefore, the pH of the solution after adding 10 mL of NaOH is 4.76.