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1) A buffer solution is prepared which is 0.210 M both in benzoic acid (C6H5CO2H) and in sodium benzoate

(C6H5CO2Na).

Ka = 6.4×10-5 for benzoic acid.


a) What is the pH of this solution?

b) When 1.20 mL of 0.0600 M HCl is added to 10.0 mL of this buffer solution, what is resulting change in pH?


2) Given that Kb = 5.6×10-4 for ethylamine:

To prepare exactly 550 mL of a buffer solution which has a pH of 10.96 and a final ethylammonium chloride salt concentration [CH3CH2NH3+] = 0.160 M, one would

need to use ________ (mass) of CH3CH2NH3Cl(s).


3) 45.0 mL of 0.100-M propenoic acid (Ka = 5.52×10-5) is titrated with 0.100-M NaOH.


a) What is the initial pH of the propenoic acid solution?

b) What is the pH of the solution after 18.0 mL NaOH has been added?

c) What is the pH of the solution after a total of 22.5 mL NaOH has been added?

d) What is the pH of the solution after a total of 36.0 mL NaOH has been added?

e) What is the pH of the solution after a total of 45.0 mL NaOH has been added?

f) What is the pH of the solution after a total of 54.0 mL NaOH has been added?


4) Consider the following two reactions involving oxalic acid.


2MnO4− + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2 + 8H2O


H2C2O4 + 2OH− → C2O42− + 2H2O


a) What volume of 0.0300 M KMnO4 solution will titrate a solution prepared from 135 mg of H2C2O4?

b) What volume of 0.0300 M NaOH solution will titrate a solution prepared from 135 mg of H2C2O4?


5) A mixed aqueous solution of HCl and sulfurous acid (H2SO3; Ka1 = 1.39×10-2; Ka2 = 6.73×10-8) is titrated with 0.1000-M NaOH.

The first end point (ethyl orange indicator) is reached after 35.16 mL of NaOH has been added.

The second end point (thymolphthalein indicator) is reached after a total of 55.72 mL of NaOH has been added.


a) What quantity of HCl (units of mol or mmol) was present in the initial mixture?

b) What quantity of sulfurous acid (units of mol or mmol) was present in the initial mixture?

Top Answer

Ans 1a- Ka = 6.4 x 10 -5 pKa = - log Ka = - log ( 6.4 x 10 -5 ) = - (log 6.4 + log 10 -5 ) = - (log 6.4 - 5 log 10) = -... View the full answer

IMG_6657.jpg

Kb = 5.6 x14
PH = 10 9 6
come of salt = 0'160 m
vat of buffer sothe 550 mad
PKbe - log kb
= - 109 5 4 x10
PKb = 3.25
= = [1095 6- 410910] = - [0 74 82 -41 = 3:25
pka = 14 - Pkb = 14- 3:25 = 10-75...

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