A 30.0 ml sample of 0.330 M propanoic acid (Ka = 1.3 x 10-5) is titrated with 0.600 M KOH. Calculate the pH after the addition of 20.5 mL of base.
Question
A 30.0 ml sample of 0.330 M propanoic acid (Ka = 1.3 x 10-5) is titrated with 0.600 M KOH. Calculate the pH after the addition of 20.5 mL of base.
Solution
Sure, here's how you can solve this problem:
Step 1: Calculate the initial moles of propanoic acid (HC3H5O2) and KOH.
Moles of HC3H5O2 = Molarity * Volume = 0.330 M * 0.030 L = 0.0099 mol Moles of KOH = Molarity * Volume = 0.600 M * 0.0205 L = 0.0123 mol
Step 2: Determine the reaction between HC3H5O2 and KOH.
HC3H5O2 + KOH -> KC3H5O2 + H2O
Step 3: Calculate the moles of HC3H5O2 and KC3H5O2 after the reaction.
Moles of HC3H5O2 = Initial moles of HC3H5O2 - moles of KOH = 0.0099 mol - 0.0123 mol = -0.0024 mol Since the moles of HC3H5O2 cannot be negative, it means all the HC3H5O2 has reacted with KOH and there is excess KOH.
Moles of KC3H5O2 = Initial moles of HC3H5O2 = 0.0099 mol
Step 4: Calculate the concentration of OH- ions from the excess KOH.
Moles of excess KOH = Moles of KOH - initial moles of HC3H5O2 = 0.0123 mol - 0.0099 mol = 0.0024 mol Volume of solution = Volume of HC3H5O2 + Volume of KOH = 0.030 L + 0.0205 L = 0.0505 L Concentration of OH- = Moles of excess KOH / Volume of solution = 0.0024 mol / 0.0505 L = 0.0475 M
Step 5: Calculate the pOH and then the pH.
pOH = -log[OH-] = -log(0.0475) = 1.32 pH = 14 - pOH = 14 - 1.32 = 12.68
So, the pH after the addition of 20.5 mL of base is approximately 12.68.
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