For the titration of 20.0 mL of 0.150 M Propionic Acid, C2H5COOH, with 0.100 M N

Question

For the titration of 20.0 mL of 0.150 M Propionic Acid, C2H5COOH, with 0.100 M NaOH, answer the following questions and construct a titration curve.

Ka- 0.000013 pKa = 4.89

A) How many milliliters of base are required to neutralize the acid (that is to reach the stoichiometric point)

B) What is the pH of the propionic acid solution BEFORE any base is added?

C) What is the pH at the half-stoichiometric point (that is, what is the pH when you have added 50% of the total volume of base required to react with all of the acid originally present?)

D) What is the pH when 95% of the base required to reach the stoichiometric point has been added?

E) What is the pH at the stoichiometric point?

F) What is the Ph when 105% of the base required to reach the stoichiometric point has been added (said another way, what is the pH when there is 5% excess of base?)

G) Graph the titration curve and label points

I know this is a long question thank you in advance for your help!!

Explanation / Answer

A)let the volume be v.so,

0.1*v=20*0.15

or v=30 ml

B)let the dissociation be x.so,

0.000013=x^2/(0.15-x)

or x=0.00139

so pH=-log(H+)

=-log(x)

=2.857

C)pH=pKa+log(salt/acid)

=pKa+log(1)

=pka

=4.89

D)pH=pKa+log(salt/acid)

=4.89 + log(0.95/0.05)

=6.17

E)pH=7+0.5pKa+0.5log(C)

=7+0.5*4.89+0.5*log(20*0.15/(20+30))

=8.83

F)[OH-]=30*0.1*0.05/(31.5+20)

=0.0029

so pH=14-pH

=11.46

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