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cvc121
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Homework Statement
a) Determine the concentration of NaCH3COO. The volume of HCl used to reach the equivalence point is 25.05 mL, the concentration of HCl is 0.10 M, and the volume of NaCH3COO solution used is 10.0 mL.
b) Using the initial concentrations of NaCH3COO and HCl, and a reliable literature value for the pKb for NaCH3COO or PKa for CH3COOH, predict the theoretical pH of the equivalence point.
Homework Equations
The Attempt at a Solution
Here is my reasoning and calculations:
a)
The equivalence point is the point in the titration when the number of moles of a standard solution (titrant) is equal to the number of moles of a solution of unknown concentration (analyte).
Therefore, at the equivalence point:
ntitrant = nanalyte
nHCl = nNaCH3COO
moles of HCl = (concentration)(volume)
moles of HCl = (0.10 M)(0.02505 L)
moles of HCl = 2.5 x 10-3 mol
moles of HCl = moles of NaCH3COO = 2.5 x 10-3 molConcentration of NaCH3COO = moles / total volume
Concentration of NaCH3COO = 2.5 x 10-3 mol / (0.01 L + 0.02505 L)
Concentration of NaCH3COO = 0.07 M
b)
A reliable literature value for the pKa for acetic acid is 4.756.
pKa = -log Ka
Ka = 10-pKa
Ka = 10-4.756
Ka = 1.754 x 10-5
At the equivalence point of the titration of a weak base with a strong acid, the general reaction H3O+ (aq) + B (aq) ---> BH+ (aq) + H2O (l) has gone approximately to completion. Therefore, approximately all sodium acetate (NaCH3COO) will be converted to its conjugate acid, acetic acid (CH3COOH), in a 1:1 molar ratio.
Ka = [CH3COO-][H3O+] / [CH3COOH]
1.754 x 10-5 = x2 / (0.07 M – x)
Ka is very small. Therefore, the x value in the denominator is negligible and can be omitted.
1.754 x 10-5 = x2 / 0.07 M
x2 = (1.754 x 10-5 )(0.07 M)
x2 = 1.228 x 10-6 M
x = 1.108 x 10-3 M
[H3O+] = x = 1.108 x 10-3 M
Validity check:
(1.108 x 10-3 M / 0.07 M )(100%) = 1.58%
Theoretical pH = -log[H3O+]
Theoretical pH = -log[1.108 x 10-3]
Theoretical pH = 2.96
Could anyone verify my reasoning and calculations? Thanks. All help is very much appreciated.