22. In order to titrate a 35.00 mL sample of acetic acid (CH3CO2H) to the equiva

Question

22. In order to titrate a 35.00 mL sample of acetic acid (CH3CO2H) to the equivalence point, 37.50 mL of a 1.870 M solution of KOH were required. Ka (CH3CO2H) = 1.8×10-5 , Kb (CH3CO2 - ) = 5.6×10-10 . The following questions refer to this titration.

a) Write a complete equation of a chemical reaction which occurs in solution during titration (before the equivalence point is reached) __________________________________________________________________

b) What was the molarity of the original CH3CO2H solution? _____________ M

c) List all the species that are present in the solution at the equivalence point. ______________________________________________________________________

d) Calculate the pH of the solution at the equivalence point? ____________________

e) What is the pH of the solution at the half-equivalence (half-way) point?

NO IDEA!!

Explanation / Answer

a)

CH3COOH +   KOH   --------------------> CH3COOK +   H2O

b)

At equivalence point :

millimoles of acid = millimoles of base

35 x M = 37.5 x 1.870

M = 2.004

molarity of the original CH3CO2H solution = 2.004 M

c)

at equivalence point

salt and water remains .

CH3COOK and H2O are present

d)

[salt] = 70.125 / (35+37.5) = 0.967 M

pH = 7 + 1/2 (pKa + log C)

    = 7 + 1/2 (4.74 + log 0.967)

pH = 9.36

e)

pH = pKa

pH = 4.74

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