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

Answer 1

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)