Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq).?
What is the pH after the addition of 15.0 mL of 0.30 M KOH(aq)?
What volume of 0.30 M KOH(aq) is required to reach halfway to the stoichiometric point?
Calculate the pH at the halfway point.
How did you get 3 mml for C6H5COOH? when calculating the halfway point
How did you get 3 mmol for C6H5COOH? when calculating the halfway point**
- HalchemistLv 79 years agoFavorite Answer
start mmol HOBz = 30.0 mL x 0.20 mmol/mL = 6.0 mmol
mmol base added = 15 mL x 0.30 mmol/mL = 4.5 mmol
mmol HObz remaining = 6.0 - 4.5 = 1.5 mmol
pH = pKa + log(OBz-/HOBz) = 4.20 + log(4.5/1.5) = 4.68
? mL KOH = 3.0 mmol HOBz x 1 mmol KOH / mmol HOBz x 1 mL / 0.30 mmol KOH
= 10. mL KOH
If Cs=Ca, then log 1 = 0 and pH = pKa = 4.20