a 250.0 ml buffer solution is 0.250 M in acetic acid and. 250M in sodium acetate. what is the ph after addition of. 0050 mol of HCL? what is the ph after the addition of. 0050 mol of NaOH?
Ka = 0.000018
To find the pH after the addition of 0.0050 mol of HCl, we need to consider the reaction between HCl and acetic acid in the buffer solution. The balanced equation for this reaction is:
HCl + CH3COOH ⇌ H3O+ + CH3COO-
Here, HCl is the strong acid (H3O+ ions) and acetic acid (CH3COOH) is the weak acid. The buffer solution helps maintain a relatively stable pH upon addition of small amounts of acid or base.
To find the pH, we need to calculate the moles of acetic acid and sodium acetate remaining in the buffer solution after the reaction.
Step 1: Calculate the moles of acetic acid initially present in the buffer:
Moles = concentration (M) * volume (L)
Moles of acetic acid = 0.250 M * 0.250 L = 0.0625 mol
Step 2: Calculate the moles of sodium acetate initially present in the buffer:
Moles of sodium acetate = 0.250 M * 0.250 L = 0.0625 mol
Step 3: Determine which reactant is limiting in the buffer solution. This will help us identify which component will react completely, leaving the other component as the excess.
Both acetic acid and sodium acetate have the same number of moles initially, so neither is limiting. However, since HCl is added, it will react with the acetic acid, and some of the acetic acid will be converted into the acetate ion.
Step 4: Calculate the moles of HCl added to the buffer:
Moles of HCl = 0.0050 mol
Step 5: Determine the moles of the remaining acetic acid after reaction:
Moles of acetic acid remaining = Moles of acetic acid initially - Moles of HCl
Moles of acetic acid remaining = 0.0625 mol - 0.0050 mol = 0.0575 mol
Step 6: Determine the moles of the conjugate base (acetate ion) formed from the reaction:
Moles of acetate ion = Moles of HCl reacted = 0.0050 mol
Step 7: Calculate the concentration of acetic acid and acetate ion in the buffer after the reaction:
Concentration of acetic acid = Moles of acetic acid remaining / volume of buffer solution
Concentration of acetic acid = 0.0575 mol / 0.250 L = 0.230 M
Concentration of acetate ion = Moles of acetate ion / volume of buffer solution
Concentration of acetate ion = 0.0050 mol / 0.250 L = 0.020 M
Step 8: Apply the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log(Acid / Conjugate base)
The pKa value for acetic acid is 4.74.
pH = 4.74 + log(0.230 M / 0.020 M)
pH = 4.74 + log(11.5)
pH = 4.74 + 1.06
pH = 5.80
Therefore, the pH after the addition of 0.0050 mol of HCl is 5.80.
To find the pH after the addition of 0.0050 mol of NaOH, we follow a similar process.
Step 1: Calculate the new moles of acetic acid after the addition of NaOH:
Moles of acetic acid remaining = initial moles of acetic acid - moles of HCl
Moles of acetic acid remaining = 0.0575 mol
Step 2: Calculate the moles of sodium acetate formed from the reaction:
Moles of sodium acetate formed = initial moles of acetate + moles of NaOH added
Moles of sodium acetate formed = 0.0625 mol + 0.0050 mol = 0.0675 mol
Step 3: Calculate the concentration of acetic acid and acetate ion in the buffer after the reaction:
Concentration of acetic acid = Moles of acetic acid remaining / volume of buffer solution
Concentration of acetic acid = 0.0575 mol / 0.250 L = 0.230 M
Concentration of acetate ion = Moles of sodium acetate formed / volume of buffer solution
Concentration of acetate ion = 0.0675 mol / 0.250 L = 0.270 M
Step 4: Apply the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log(acid / conjugate base)
pH = 4.74 + log(0.230 M / 0.270 M)
pH = 4.74 - log(0.85)
pH = 4.74 - (-0.07)
pH = 4.74 + 0.07
pH = 4.81
Therefore, the pH after the addition of 0.0050 mol of NaOH is 4.81.
To find the pH of the buffer solution after the addition of acids or bases, we need to consider the Henderson-Hasselbalch equation:
pH = pKa + log [salt]/[acid]
Here, the acetic acid is the acid and sodium acetate is the salt.
1. pH after addition of 0.0050 mol HCl:
First, we need to determine the concentration of acetic acid and sodium acetate after the addition of HCl.
- Acetic acid concentration:
Initial moles of acetic acid = Volume x Concentration
Initial moles of acetic acid = 0.250 L x 0.250 M
Initial moles of acetic acid = 0.0625 mol
Final moles of acetic acid = Initial moles of acetic acid - moles of HCl
Final moles of acetic acid = 0.0625 mol - 0.0050 mol
Final moles of acetic acid = 0.0575 mol
- Sodium acetate concentration:
Initial moles of sodium acetate = Volume x Concentration
Initial moles of sodium acetate = 0.250 L x 0.250 M
Initial moles of sodium acetate = 0.0625 mol
Since HCl does not react with sodium acetate, the moles of sodium acetate remain the same.
Now, we can calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log [salt]/[acid]
pH = pKa + log (0.0625 mol/0.0575 mol)
Alternatively, we can calculate the ratio [salt]/[acid]:
[salt]/[acid] = (0.0625 mol)/(0.0575 mol)
[salt]/[acid] = 1.087
Therefore, pH = pKa + log 1.087
2. pH after addition of 0.0050 mol NaOH:
We follow the same procedure as above to determine the concentration of acetic acid and sodium acetate after the addition of NaOH.
- Acetic acid concentration:
Initial moles of acetic acid = Volume x Concentration
Initial moles of acetic acid = 0.250 L x 0.250 M
Initial moles of acetic acid = 0.0625 mol
Since NaOH reacts with acetic acid to form water and sodium acetate:
Moles of acetic acid left = Initial moles of acetic acid - moles of NaOH
Moles of acetic acid left = 0.0625 mol - 0.0050 mol
Moles of acetic acid left = 0.0575 mol
- Sodium acetate concentration:
Initial moles of sodium acetate = Volume x Concentration
Initial moles of sodium acetate = 0.250 L x 0.250 M
Initial moles of sodium acetate = 0.0625 mol
Final moles of sodium acetate = Initial moles of sodium acetate + moles of NaOH
Final moles of sodium acetate = 0.0625 mol + 0.0050 mol
Final moles of sodium acetate = 0.0675 mol
Now, we can calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log [salt]/[acid]
pH = pKa + log (0.0675 mol/0.0575 mol)
Alternatively, we can calculate the ratio [salt]/[acid]:
[salt]/[acid] = (0.0675 mol)/(0.0575 mol)
[salt]/[acid] = 1.174
Therefore, pH = pKa + log 1.174
Note: To calculate the final pH, you will need the pKa value for acetic acid.
250 mL x 0.250M HAc(CH3COOH) = 62.5 mmoles to begin.
250 mL x 0.250M NaAc(CH3COONa) = 62.5 mmoles to begin. Write the reaction and make a table. It proves invaluable.
............Ac^- + H^+ ==> HAc
begin......62.5.....0.......62.5
add H^+.............5.........
react......-5.0....-5.......+5.0
final......57.5.....0........67.5
pH = pKa + log [(57.5)/(67.5)] = ??
For the addition of 0.0050 mol (5.0 mmols) base, I do this.
.............HAc + OH^- ==> Ac^- + H2O
begin........62.5...0.......62.5....0
add OH^-............5.0................
react........-5.0..-5.0......+5.0.+5.0
final........57.5...0........67.5...5.0
pH = pKa + log[(67.5/57.5) = ??