The next 9 questions are related to the titration of 30.00 mL of a 0.0750 M acet

ID: 968093 • Letter: T

Question

The next 9 questions are related to the titration of 30.00 mL of a 0.0750 M acetic acid solution with 0.0800 M KOH.

What is the initial pH of the analyte solution?

What volume of KOH is required to reach the equivalence point of the titration (in mL)?

How many mmol of the salt are present at the equivalence point? (ANALYTICAL AMOUNT, NOT EQUILIBRIUM AMOUNT)

What is the volume of the solution at the equivalence point (in mL)?

What is the molar concentration of the salt at the equivalence point? (ANALYTICAL CONCENTRATION, NOT EQUILIBRIUM CONCENTRATION)

What is the pOH at the equivalence point?

What is the pH at the equivalence point?

How many mL of KOH have been added at the half-equivalence point?

What is the pH at the half-equivalence point?

Remember that the acetate ions react with water. This is the reaction of a base with water so you need to use the Kb expression.

No of mol of CH3COOH = 30/1000*0.075 = 0.00225 mol

at equivalence point , No of mol of CH3COOH = No of mol of KOH

No of mol of KOH = 0.00225 mol

volume of KOH = 0.00225/0.08 = 28.125 ml

mmol of the salt are present at the equivalence point

= 28.125*0.08 = 2.25 mmol

volume of the solution at the equivalence point = 28.125+30 = 58.125 ml

at equivalence point

concentration of salt = n/v = 2.25/58.125 = 0.039 M

pH = 7+1/2(pka+logC)
= 7+1/2(4.74+log0.039)

= 8.66

pOH = 14 - 8.66 = 5.34

pH at the half-equivalence point = pka = 4.74

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