Part A a solution that is 6.5×10 2 M in potassium propionate (C2H5COOK or KC3H5O

Question

Part A

a solution that is 6.5×102 M in potassium propionate (C2H5COOK or KC3H5O2 ) and 9.0×102 M in propionic acid (C2H5COOH or HC3H5O2 )

Express your answer using two significant figures.

Part B

a solution that is 7.0×102 M in trimethylamine, (CH3)3N, and 0.13 M in trimethylammonium chloride, (CH3)3NHCl

Express your answer using two significant figures.

Part C

a solution that is made by mixing 50.0 mL of 0.14 M acetic acid and 50.0 mL of 0.22 M sodium acetate

Express your answer using two significant figures.

Explanation / Answer

PART A) using the Henderson - Hasselbalch equation:  

pKa propionic acid = 4.87 (from internet)

Molarity of CH3CH2COOH = 9E-2M

Molarity of CH3CH2COOK = 6.5E-2M

Equation:

pH = pKa + log ([salt]/[acid] )

pH = 4.87 + log (6.5E-2/9E-2)

pH = 4.87 + log 0.722

pH = 4.87 + (-0.14)

pH = 4.728 = 4.73

PART B) pKb for (CH3)3N = 4.19 (from data)

pOH = pKa + log (salt]/[base])

pOH = 4.19 + log ( 0.13/7E-2)

pOH = 4.19 + log 1.85

pOH = 4.19 + 0.268

pOH = 4.4588

pH = 14.00- pOH

pH = 9.54

PART C)pKa for CH3COOH = 4.74 (from data)

Using Molarity of acetic acid: M1V1 = M2V2

M1x100 = 0.14 x 50

M1 = 0.14x50/100 = 0.07M

Also,Molarity of CH3COONa

M1*100 = 0.22x50/100

M1= 0.11M

Now pH = pKa + log ([salt]/[acid])

pH = 4.74 + log( 0.11/0.07)

pH = 4.74 + log 1.57

pH = 4.74 + 0.19

pH = 5.056 = 5.06

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