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# Titration of 50 mL 0.108 M propanoic acid (HPr) with 0.150 M NaOH

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A sample of 0.0054 mol propanoic acid was dissolved in water to give 50ml solution. This solution is titrated with 0.150 M of NaOH. Draw the expected titration curve labeling the equivalence point (including the volume of NaOH added and the pH) and show the pKa point (labeling the pH value and volume of NaOH added)

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This answer provides a full solution to the question. As well, it provides an example for understanding weak acid-strong base titration cases, an illustration of a titration curve demonstrating this and outlines why understanding this case may help solve the problem. 500 words.

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Titration of 50 mL 0.108 M propanoic acid (HPr) with 0.150 M NaOH

Calculate the pH at any stage of the titration:
a) Original solution of HPr
HPr (aq) = H+ (aq) + Pr (aq)
[init] 0.108 M 0 0
[change] -x +x +x
[equil] (0.108-x) x x

Ka = x^2/(0.108 -x)

x = = 1.22 x 10^-3 M = [H3O+]

So pH = 2.91

In the buffer region (see graph) at the pKa point, pKa = pH, because [Hpr] = [Pr^-]
Volume of NaOH added at pH=4.86 can be read from the graph if plotted properly.
Use Henderson-Hasselbalch equation and you don't have to calculate new [HPr] and [Pr ]; can use ratio of n(HPr) and n(Pr ).
pH = pKa + log [Pr-]/[HPr]
Vf's cancel pH = pKa + log (moles ...

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