Here is a series of three acid-base equilibrium problems. They are designed to d
ID: 717579 • Letter: H
Question
Here is a series of three acid-base equilibrium problems. They are designed to demonstrate how the solution procedure is similar even when the unknown parameter changes. a) What is the pH of a solution prepared by adding 2.7x 104 moles of benzoic acid (pKa 4.20) to b) How many moles of sodium acetate (the pKa for acetic acid is 4.75) should be added to 1.0 L of c) If I 045 moles of an acid were added to 1.0 L of water and the equilibrium pH = 6.3, what is the 1.0 L of water? water to adjust the pH-7.1? pKa of the acid?Explanation / Answer
a) For the dissociation reaction:
C7H6O2 + H2O = C7H5O2- + H3O +
where in the equilibrium the concentrations are:
[C7H6O2] = 2.7x10 ^ -4 - x
[C7H5O2-] = x
[H3O +] = x
Ka = 10 ^ -pKa = 10 ^ -4.20 = 6.31x10 ^ -5
Ka = [C7H5O2-] * [H3O +] / [C7H6O2] = x ^ 2 / 2.7x10 ^ -4 - x = 6.31x10 ^ -5
By clearing and solving equations of the second degree we have:
x = 1.03x10 ^ -4 M = [H3O +]
and we calculate pH = - Log [H3O +] = - Log (1.03x10 ^ -4 M) = 4.
b) Starting from the reaction:
CH3COONa + H2O = CH3COOH + NaOH
in equilibrium the concentrations are:
[CH3COONa] = a - x
[CH3COOH] = x
[NaOH] = x
Ka = 10 ^ -pKa = 10 ^ -4.75 = 1.78x10 ^ -5
Kb = Kw / Ka = 10 ^ -14 / 1.78x10 ^ -5 = 5.62x10 ^ -10
so:
Kb = [CH3COOH] * [NaOH] / [CH3COONa] = x ^ 2 / a-x
pOH = 14 - pH = 14 - 7.1 = 6.9
x = [OH-] = 10 ^ -pOH = 1.26x10 ^ -7
clearing to (moles of salt needed to add):
a = (x ^ 2 / Kb) + x = 2.84x10 ^ -5 moles of salt needed to add.
c) Assuming reaction:
HA + H2O = A- + H3O +
where in the equilibrium the concentrations are:
[A-] = [H3O +] = x
[HA] = 10 ^ -4.5 - x
pH = 6.3
[H3O +] = 10 ^ -6.3 = 5.01x10 ^ -7 = x
so:
Ka = x2 / 10 ^ -4.5 - x = (5.01x10 ^ -7) ^ 2/10 ^ -4.5 - 5.01x10 ^ -7 = 8.07x10 ^ -9
pKa = - Log Ka = - Log (8.07x10 ^ -9) = 8.09.
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