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 w

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?     Ka   for propionic acid is    1.3×10–5   at  25°C.Ka   for propionic acid is    1.3×10–5   at  25°C.

Select one:

a. 8.93

b. 7.65

c. 9.47

d. 5.98

e. 9.11

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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