A 35.0 mL sample of 0.150 M acetic acid is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added:
a) 0mL, b) 17.5 mL, c) 34.5 mL, d) 35 mL
I've figured out a,b, and c. But at 35 mL the moles become equal so when I plug the numbers into Henderson Hasselbach equation I get log to be 1 which makes the pH be 4.74 which is obviously not right. What am I doing wrong?
I don't think you can use the HH equation for the equivalence point. You have the base (acetate concn) but there is no acid and a number/0 is undefined.
CH3COOH + NaOH ==> CH3COONa + H2O
What you do is the pH at the equivalence point is due to the hydrolysis of th salt.
CH3COO^- + HOH ==> CH3COOH + OH^-
Kb = (Kw/Ka) = (CH3COOH)(OH^-)/(CH3COO^-)
You let x = CH3COOH and x = OH^-, you know Kw, Ka, and acetate, solve for x and convert OH^- to pH.
35.0 mL sample of 0.150 M acetic acid (HC2H3O2) (Ka = 1.8 x 10-5) is titrated with 0.150 M NaOH solution. Calculate the pH after 17.5 mL volumes of base have been added
47 mL
To calculate the pH at the equivalence point, when the moles of acetic acid and NaOH are equal, you need to consider the species present in solution and their respective concentrations.
At the equivalence point, all the acetic acid (CH3COOH) is converted to its conjugate base, acetate ion (CH3COO-), and the same number of moles of NaOH are consumed to produce the same number of hydroxide ions (OH-) in solution.
However, the pH can be affected by the presence of excess NaOH. So, let's go through the steps to calculate the pH at 35 mL:
Step 1: Calculate the moles of acetic acid initially present.
Molarity (M) = Moles (mol) / Volume (L)
0.150 M = Moles / 0.035 L
Moles of acetic acid = 0.150 M x 0.035 L = 0.00525 mol
Step 2: Calculate the moles of NaOH required to reach the equivalence point.
Since the reaction between acetic acid and NaOH is 1:1, the moles of NaOH needed would be the same as the moles of acetic acid.
Moles of NaOH = 0.00525 mol
Step 3: Calculate the volume of NaOH solution required to reach the equivalence point.
Molarity (M) = Moles (mol) / Volume (L)
0.150 M = 0.00525 mol / Volume (L)
Volume of NaOH solution = 0.00525 mol / 0.150 M = 0.035 L = 35 mL
So, at 35 mL, you have reached the equivalence point. However, the pH cannot be accurately determined using the Henderson Hasselbalch equation because the acetic acid has been completely neutralized by NaOH, and the concentration of acetic acid or its conjugate base (acetate ion) is negligible.
At the equivalence point, the pH depends on the autoionization of water, which can be assumed to be 7 in this case. This means that the pH at 35 mL is 7, because the solution is neutral.