Consider the titration of a 23.0 mL sample of 0.105 molL1 CH3COOH ( K a=1.8×105)

Question

Consider the titration of a 23.0 mL sample of 0.105 molL1 CH3COOH (Ka=1.8×105) with 0.125 molL1 NaOH. Determine each quantity:

Part A

Part complete

the initial pH

Express your answer using two decimal places.

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Part B

the volume of added base required to reach the equivalence point

20.1620.16

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Part C

the pH at 6.0 mL of added base

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Part D

the pH at one-half of the equivalence point

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Part E

the pH at the equivalence point

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Explanation / Answer

CH3COOH <-- --> CH3COO- + H+
initial 0.105 0 0
equilibrium 0.105-x x x

(a) Ka = 1.8 x 10-5 = x^2/ 0.105-x

x = [H+] =0.0013658 M
pH = -log[H+] = 2.86

(b)

Writing volume in L to get the no. of moles
moles acetic acid = 0.0230 x 0.105=0.002415
moles NaOH required = 0.002415
Volume NaOH = 0.002415/ 0.125 M=0.01932 L => 19.32 mL

(c)
moles NaOH = 6.00 x 10^-3 L x 0.125 M=0.00075
moles acetic acid in excess = 0.002415 - 0.00075=0.001665
moles acetate formed = 0.00075
pKa for acetic acid = 4.74
Using Buffer eqn, pH = 4.74 + log 0.00075/ 0.001665=4.39

(d)
At one half equivalence point moles acid = moles acetate, thus log term vanishes in buffer equation.
pH = 4.74

(e)
volume NaOH = 0.01932 L
total volume = 0.0178+ 0.0230 =0.04232 L
moles acetate formed = 0.002415
concentration acetate = 0.002415/ 0.04232 =0.0570 M

Then,

CH3COO- + H2O <-- --> CH3COOH + OH-
0.057 0 0
0.057-x x x

Kb = Kw/Ka =10-14 / 1.8 x 10-5 = 5.56 x 10^-10  = x2 / 0.057-x
x = [OH-]=5.63 x 10-6 M
pOH = 5.25
pH = 8.75

Thank You

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