A titration involves adding a reactant of known quantity to a
solution of an another reactant while monitoring the equilibrium
concentrations. This allows one to determine the concentration of
the second reactant. The equation for the reaction of a generic
weak acid HA with a strong base is
HA(aq)+OH−(aq)→A−(aq)+H2O(l) |
A certain weak acid, HA, with a Ka value of 5.61×10−6, is titrated with NaOH. Part A A solution is made by titrating 9.00 mmol (millimoles) of HA and 2.00 mmol of the strong base. What is the resulting pH? Express the pH numerically to two decimal places.
Part B More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 43.0 mL ? Express the pH numerically to two decimal places.
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HA + OH- = A- + H2O
mmol HA = 9.00 - 2.00 = 7.00
mmol A- = 2.00
pKa = - log Ka
Ka = 5.61×10−6
So
Pka = -log Ka = -log (5.61×10−6)= 5.25
According to Henderson –hasselbach equation
p H = p K a + l o g ( [ A − ] [ H A ]
pH = 5.25 + log 2.00/7.00=4.71
Part
B
1mmol = 0.001 mol
Volume = 43 ml
[A-] = 0.0090 mol/ 0.0430 L=0.209 M
A- + H2O <-----> HA + OH-
Kb = Kw/Ka = 1.00 x 10‑14/ 5.61×10−6 = 1.78 x 10
-10
1.78 x 10-10 = x2/ 0.209-x
1.78 x 10-10 x 0.209 = x2
X2 = 3.7 x 10-11
therefore
x = 6.1 x 10-6M
x = [OH-]= 6.1 x 10-6M
pOH =-log [6.1 x 10-6M]= 5.21
pH = 14 – 5.21=8.79
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