a. How many moles of the weak acid were added to the beaker? b. How many moles o
ID: 793896 • Letter: A
Question
a. How many moles of the weak acid were added to the beaker?
b. How many moles of NaOH are required to neutralize the sample of weak acid?
c. how many mililiters of the NaOH are required to neutralize the sample of weak acid?
d. How many moles of NaOH have been added at one half of the volume in part 'c' (volume at half equivalence point)?
e. How many moles of the weak acid have reacted at the half equivalence point?
f. Calculate the pH of the solution at the half equivalence point.
g. Explain how the pH at the half equivalence point is related to Ka for the weak acid.
Explanation / Answer
a)
No. of moles of HF added to beaker= V*M=(20/1000)*0.33=6.6 *10^-3 moles
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b)
No.of moles of NaOH= No. of moles of HF
No. of moles of NaOH required to neutralize the sample of weak acid =6.6 *10^-3 moles
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c)
No. of millilitres of NaOH required=No. of millimoles/M
=6.6/0.11= 60 mL
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d)
No. of moles of NaOH required at half-equivalence point =Total moles/2=
=3.3 *10^-3 moles
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e)
No. of moles of NaOH reacted at half-equivalence point =3.3 *10^-3 moles
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f)
pKa of HF=3.2
Ka=[H+][F-]/[HF]
Ka=[H+]^2/[HF]
Taking negative log on both sides,
pka=2pH + log[HF]
At half-equivalence,
No. of moles of HF=3.3 *10^-3 moles
Total volume=20 + 30 =50 mL
[HF]=3.3/50=0.066
pH=(3.2 - log 0.66)/2
pH=2.19
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g)
Since, weak acid never dissociates completely. At the equivalence point, th amount of H+ and consequently the pH is determined by the dissociation constant Ka of weak acid.
Ka=[H+][F-]/[HF]
Ka=[H+]^2/[HF]
Taking negative log on both sides,
pka=2pH + log[HF]
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