Calculate the ph during the titration of 20.0ml 0.5000M ethanoic acid Ka=0.0000175 with 0.500M NaOH after the addition of 0.0ml and 10.0ml NaOH

addition of 0.0ml and 10.0ml NaOH >> ????????

To calculate the pH during the titration of 20.0 mL of 0.5000 M ethanoic acid (Ka = 0.0000175) with 0.500 M NaOH after the addition of 0.0 mL and 10.0 mL NaOH, we need to use the concept of acid-base reactions and the Henderson-Hasselbalch equation.

Step 1: Calculate the initial moles of ethanoic acid (acetic acid):
Moles of ethanoic acid = volume (in L) x molarity = 0.0200 L x 0.5000 mol/L = 0.0100 mol

Step 2: Determine the amount of NaOH added:
For 0.0 mL NaOH: 0.0 mL NaOH corresponds to 0.0 L NaOH
For 10.0 mL NaOH: 10.0 mL NaOH corresponds to 0.0100 L NaOH

Step 3: Determine the moles of NaOH added:
Moles of NaOH = volume (in L) x molarity
For 0.0 mL NaOH: Moles of NaOH = 0.0 L x 0.500 mol/L = 0.0000 mol
For 10.0 mL NaOH: Moles of NaOH = 0.0100 L x 0.500 mol/L = 0.0050 mol

Step 4: Calculate the final moles of ethanoic acid remaining after each addition of NaOH:
For 0.0 mL NaOH: Final moles of ethanoic acid = initial moles - moles of NaOH added = 0.0100 mol - 0.0000 mol = 0.0100 mol
For 10.0 mL NaOH: Final moles of ethanoic acid = initial moles - moles of NaOH added = 0.0100 mol - 0.0050 mol = 0.0050 mol

Step 5: Use the Henderson-Hasselbalch equation to calculate the pH:

pH = pKa + log ([A-] / [HA])

Where:
pH = concentration of hydrogen ions (H+)
pKa = -log Ka (Ka is the acid dissociation constant)
[A-] = concentration of the acetate ion (ethanoate) after the addition of NaOH
[HA] = concentration of the ethanoic acid

For each addition of NaOH, we will calculate the pH using the Henderson-Hasselbalch equation.

For 0.0 mL NaOH:
[A-] = final moles of ethanoic acid = 0.0100 mol
pH = pKa + log ([A-] / [HA]) = -log (0.0000175) + log (0.0100 / 0.0100) = -log (0.0000175) = 4.76

For 10.0 mL NaOH:
[A-] = final moles of ethanoic acid = 0.0050 mol
pH = pKa + log ([A-] / [HA]) = -log (0.0000175) + log (0.0050 / 0.0100) = -log (0.0000175) - 0.301 = 4.76 - 0.301 = 4.46

Thus, the pH after the addition of 0.0 mL NaOH is 4.76, and the pH after the addition of 10.0 mL NaOH is 4.46.