5. You want to determine the acid constant for propanoic acid, C2H5COOH. 3.15 g
ID: 580426 • Letter: 5
Question
5. You want to determine the acid constant for propanoic acid, C2H5COOH. 3.15 g of the acid was dissolved in water and diluted in a graduated flask to 100 cm 3. The pH of the solution was 4.13. Which of the following answers is the correct acid constant for this example? EXPLAIN ALSO WHY.
a.Ka = 1,8 * 10-2 mol/dm3
b.Ka = 1,3 * 10-7 mol/dm3
c.Ka = 1,3 * 10-5 mol/dm3
d.Ka = 1,8 * 10-5 mol/dm3
5. You want to determine the acid constant for propanoic acid, C2H5COOH. 3.15 g of the acid was dissolved in water and diluted in a graduated flask to 100 cm 3. The pH of the solution was 4.13. Which of the following answers is the correct acid constant for this example? EXPLAIN ALSO WHY.
a.Ka = 1,8 * 10-2 mol/dm3
b.Ka = 1,3 * 10-7 mol/dm3
c.Ka = 1,3 * 10-5 mol/dm3
d.Ka = 1,8 * 10-5 mol/dm3
5. You want to determine the acid constant for propanoic acid, C2H5COOH. 3.15 g of the acid was dissolved in water and diluted in a graduated flask to 100 cm 3. The pH of the solution was 4.13. Which of the following answers is the correct acid constant for this example? EXPLAIN ALSO WHY.
a.Ka = 1,8 * 10-2 mol/dm3
b.Ka = 1,3 * 10-7 mol/dm3
c.Ka = 1,3 * 10-5 mol/dm3
d.Ka = 1,8 * 10-5 mol/dm3
Explanation / Answer
Molar mass of C2H5COOH = 74.07 g/mol
Moles of C2H5COOH = 3.15 g / 74.07 g/mol => 0.0425 moles
[C2H5COOH] = 0.0425 moles / 100 dm3
[C2H5COOH] = 0.000425 mol/dm3
pH = -log[H+]
[H+] = 10-pH => 10-4.13
[H+] = 0.0000741 mol/dm3
CH3CH2COOH <====> H+ (aq) + CH3CH2COO- (aq)
Ka = [H+] [CH3CH2COO-]/[CH3CH2COOH]
[H+] = [CH3CH2COO-]
Ka = [H+]2 / [CH3COOH]
Ka = (0.0000741 mol/dm3)2/ 0.000425 mol/dm3
Ka = 0.0000129 mol/dm3
Ka = 1.29 x 10-5 mol/dm3
So Answer : 1.3 x 10-5 mol/dm3 (option C)
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