A 25.00-mL sample of propionic acid, HC3H5O2, of unknown concentration A 25.00-m

Question

A  25.00-mL  sample of propionic acid,  HC3H5O2,  of unknown concentration A  25.00-mL  sample of propionic acid,  HC3H5O2,  of unknown concentration  was titrated with  0.151  M  KOH. The equivalence point was reached when was titrated with  0.151  M  KOH. The equivalence point was reached when  41.28  mL  of base had been added. What is the pH at the equivalence point? 41.28  mL  of base had been added. What is the pH at the equivalence point? Kafor propionic acid is 1.3×10–5at  25°C.Kafor propionic acid is 1.3×10–5at  25°C.

Select one:

a. 8.93

b. 7.65

c. 9.47

d. 5.98

e. 9.11

Explanation / Answer

Lets denote the acid by HA.

At equivalence point, the following reaction takes place:

A- + H2O ---> HA + OH-

Initial c 0 0

Eqb c-x x x

For this reaction we have:

Kb = x2/(c-x) = 10-14/Ka = (10-14)/(1.3*10-5)

Moles of A- present initially = Moles of KOH used = Molarity*Volume = 0.151*0.04128 = 0.00623

So,

c = Moles/Volume = 0.00623/(0.025+0.04128) = 0.0939 M

Putting values we get:

x2/(0.0939-x) = (10-14)/(1.3*10-5)

Solving we get:

x = 8.49*10-6

So,

[OH-] = x = 8.49*10-6 M

So,

pOH = -log([OH-]) = 6-log(8.49) = 5.07

So,

pH = 14-pOH = 14-5.07 = 8.93

Hope this helps !

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