Question

In a 100.0 mL sample of a different buffer solution, the propanoic acid solution is 0.350 M and the propanoate concentration is 0.809 M. 0.00400 moles of solid KOH are added to the 100.0 mL sample of buffer solution mentioned. Assuming that the KOH does not change the volume of the solution, what is the new pH?

Answer 1

For Propionic acid/Propionate buffer given that,

[Propanoic acid] = 0.350 M, [Propanoate] = 0.809 M, pKa = 4.87 for propionic acid.

KOH is a strong base and dissociates completely and hence [OH^-] = [KOH] =0.004 M

After addition of KOH base obviously concentration of propionic acid decreases and that of conjugate base propionate ion increases. Let us find out new concentrations,

The volume of buffer solution is 100 mL = 0.1 L

New [Propanoic acid] = original [propanonic acid] x 0.1 L – [KOH]

                                          = 0.350 x 0.1 – 0.004

                                         =0.0350 – 0.004

                                          = 0.0310 M

New [Propanoate] = 0.809 x 0.1 + 0.004 = 0.0849 M

Now let us find out pH of buffer with these new concentrations,

Henderson equation,

pH = pKa + log{[Propanoic acid]/[Propanoate]}

pH = 4.87 + log(0.0310/0.0849)

pH = 4.87 + log(0.3651)

pH = 4.87 + (-0.4375)

pH = 4.432

Hence new pH = 4.432.

Note : Original pH not calculated as not needed for calculating new pH here. pKa for Propanoic acid taken from pH data literature.

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