When a 13.62 mL volume of vinegar is used in a titration, it is found that 27.93
ID: 810869 • Letter: W
Question
When a 13.62 mL volume of vinegar is used in a titration, it is found that 27.93 mL of 0.150 M NaOH is needed to reach the equivalence point.
A. Calculate the moles of NaOH added in this titration.
B. Calculate the moles of CH3COOH in the vinegar sample.
C. Calculate the molarity of CH3COOH in the vinegar solution.
D. Calculate the moles of of CH3COONa produced in this reaction.
E. Calculate the molarity of the CH3COONa produced in this titration before any CH3COO- reacts with water.
F. Calculate the molarity of CH3COO- produced in this titration, assuming that CH3COONa dissociates completely in water, but before any CH3COO- reacts with water.
G. Calculate the pH of this titrated solution at the equivalence point.
Explanation / Answer
Given, Volume of NaOH = 27.93 mL = 0.0279 L
Molarity of NaOH = 0.150 M
A.
Moles of NaOH = Molarity x Volume(inL) = 0.150 x 0.0279 = 4.2 x 10^-3 moles
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B.
The chemical reaction is
CH3COOH + NaOH -----------> CH3COONa + H2O
As the two reactants react in equimolar ratio, moles of NaOH = Moles of CH3COOH in vinegar
Thus, moles of CH3COOH in vinegar sample = 4.2 x 10^-3 moles
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C.
Moles of CH3COOH in vinegar sample = 4.2 x 10^-3 moles
Volume of vinegar sample = 13.62 mL = 0.01362 L
Molarity of CH3COOH in vinegar = moles / Volume = 4.2 x 10^-3 moles / 0.01362 = 0.308 M
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D.
Total volume in the reaction vessel = 13.62 mL vinegar + 27.93 mL NaOH = 41.55 mL = 0.04155 L
[CH3COOH] = 4.2 x 10^-3 moles / 0.04155 L = 0.1011 M
[NaOH] = 4.2 x 10^-3 moles / 0.04155 L = 0.1011 M
The chemical reaction is
CH3COOH + NaOH -----------> CH3COONa + H2O
I 0.1011 0.1011 0 0
C -x -x +x +x
E 0.1011-x 0.1011-x +x +x
Ka = x^2 / (0.1011-x)^2 = 1.77 x 10^-5
x^2 = 1.78 x 10^-6
x = 1.34 x 10^-3 M
Thus, [CH3COONa] = 1.34 x 10^-3 M
Moles of CH3COONa = 1.34 x 10^-3 M x 0.04155 L = 5.55 x 10^-5 mol
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E.
From the calculations carried out in D, [CH3COONa] = 1.34 x 10^-3 M
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F.
[CH3COONa] = 1.34 x 10^-3 M
[CH3COO-] = 1.34 x 10^-3 M
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G.
[H+] = x = 1.34 x 10^-3 M
pH = -log[H+]
pH = -log[ 1.34 x 10^-3]
pH = 2.87
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