35.0mL sample of 0.150M acetic acid9HC2H3O2)(Ka=1.8*10-5) is titrated with 0.150M NaOH solution. calculate the pH AFTER 17.5 Ml volumes of base have been added.
To calculate the pH after adding 17.5 mL of NaOH solution, we need to determine the moles of acetic acid and NaOH that have reacted, and then calculate the concentration of acetic acid and its conjugate base.
Let's begin step-by-step.
Step 1: Determine the moles of acetic acid (HC2H3O2) in the sample.
Moles of acetic acid (HC2H3O2) = volume (in L) x molarity
= 0.035 L x 0.150 M
= 0.00525 moles
Step 2: Determine the moles of NaOH that have reacted.
Since the balanced chemical equation for the reaction between acetic acid and NaOH is 1:1, the moles of NaOH that have reacted will be equal to 0.00525 moles.
Step 3: Determine the remaining moles of acetic acid.
Since the moles of acetic acid and NaOH that have reacted are the same, the remaining moles of acetic acid would be 0.00525 moles - 0.00525 moles = 0 moles.
Step 4: Determine the new volume of the solution after adding 17.5 mL of NaOH.
New volume = initial volume + volume of NaOH added
= 35.0 mL + 17.5 mL
= 52.5 mL
= 0.0525 L
Step 5: Determine the new concentration of acetic acid.
New concentration of acetic acid = moles of acetic acid / new volume
= 0 moles / 0.0525 L
= 0 M (since all the acetic acid has reacted)
The concentration of acetic acid after adding 17.5 mL of NaOH is 0 M.
Step 6: Determine the concentration of the acetate ion (C2H3O2-) which is the conjugate base of acetic acid.
Since the reaction is complete and all the acetic acid has reacted, the moles of acetate ion formed equal the moles of NaOH added (0.00525 moles).
Concentration of acetate ion = moles of acetate ion / new volume
= 0.00525 moles / 0.0525 L
= 0.1 M
The concentration of the acetate ion after adding 17.5 mL of NaOH is 0.1 M.
Step 7: Calculate the pH.
To calculate the pH, we need to use the equilibrium expression for the dissociation of acetic acid:
Ka = [C2H3O2-][H3O+]/[HC2H3O2]
Since the concentration of acetic acid is now 0 M, we can ignore it in the equilibrium expression:
Ka = [C2H3O2-][H3O+]
Now, solve for [H3O+] to find the pH:
[H3O+] = Ka / [C2H3O2-]
= (1.8 × 10^-5) / (0.1)
= 1.8 × 10^-4
pH = -log[H3O+]
= -log(1.8 × 10^-4)
= 3.74
Therefore, the pH after adding 17.5 mL of NaOH is approximately 3.74.
To calculate the pH after 17.5 mL of NaOH solution have been added, we need to determine how many moles of acetic acid and moles of NaOH have reacted and find the remaining moles of acetic acid in the solution. From there, we can calculate the concentration of acetic acid remaining and use that information to determine the pH.
1. Calculate the number of moles of acetic acid initially present:
moles of acetic acid = volume (in L) x molarity
moles of acetic acid = 0.035 L x 0.150 mol/L = 0.00525 mol
2. Calculate the number of moles of NaOH:
moles of NaOH = volume (in L) x molarity
moles of NaOH = 0.0175 L x 0.150 mol/L = 0.002625 mol
3. Determine the limiting reactant:
The balanced equation for the reaction between acetic acid (HA) and NaOH is:
HA + NaOH → H2O + NaA
Since the stoichiometric ratio between acetic acid and NaOH is 1:1, both react completely. Therefore, acetic acid is the limiting reactant.
4. Calculate the moles of acetic acid remaining:
moles of acetic acid remaining = initial moles - moles of NaOH reacted
moles of acetic acid remaining = 0.00525 mol - 0.002625 mol = 0.002625 mol
5. Calculate the volume of solution after 17.5 mL of NaOH have been added:
volume of solution = initial volume + volume NaOH added
volume of solution = 0.035 L + 0.0175 L = 0.0525 L
6. Calculate the concentration of acetic acid:
concentration of acetic acid = moles of acetic acid remaining / volume of solution
concentration of acetic acid = 0.002625 mol / 0.0525 L = 0.05 mol/L
7. Calculate the pOH of the acetic acid solution:
pOH = -log10(concentration of acetic acid)
pOH = -log10(0.05) = 1.3
8. Calculate the pH of the acetic acid solution:
pH = 14 - pOH
pH = 14 - 1.3 = 12.7
Therefore, the pH of the solution after 17.5 mL of NaOH have been added is approximately 12.7.