The secret to this type problem is to know where you are on the titration curve.
C6H5COOH + KOH ==> C6H5COOK + H2O
For (a),
C6H5COOH + H2O ==> C6H5COO^- + H3O^+
Set up an ICE chart, substitute into Ka expression, and solve for H3O^+, then convert to pH.
For the others.
b,c,d. Calculate moles C6H5COOH initially. Calculate moles KOH added. Subtract moles KOH from moles benzoic acid to arrive at the excess benzoic acid, then substitute into either (1)Ka expression or (2) Henderson-Hasselbalch equation.
e. You find the stoichiometry point by mLa x Ma = mLb x Mb
a = acid; b = base.
f. You are at the equivalence point. The pH is determined by the hydrolysis of the salt, C6H5COOK (C6H5COO^- acting as a base).
C6H5COO^- + HOH ==> C6H5COOH + OH^-
Kb = (Kw/Ka) = (C6H5COOH)(OH^-)/(C6H5COO^-)
Post all of your work if you get stuck.
Set up ICE chart and solve for OH^- and convert to pH.
Suppose that 30.0 mL of 0.20 M C6H5COOH(aq) is titrated with 0.30 M KOH(aq).
(a) What is the initial pH of the 0.20 M C6H5COOH(aq)?
(b) What is the pH after the addition of 15.0 mL of 0.30 M KOH(aq)?
(c) What volume of 0.30 M KOH(aq) is required to reach halfway to the stoichiometric point?
mL
(d) Calculate the pH at the half stoichiometric point.
(e) What volume of 0.30 M KOH(aq) is required to reach the stoichiometric point?
mL
(f) Calculate the pH at the stoichiometric point.
Please show how to do it not just the answers.
2 answers
im still a bit confused about part c
i calculated:
n(C6H5COOH)initial= .006 mol
but when i calculate the KOH do i do the initial or with the added 15mL?
im a bit confused about that
i calculated:
n(C6H5COOH)initial= .006 mol
but when i calculate the KOH do i do the initial or with the added 15mL?
im a bit confused about that
2 answers