Question

Acetylsalicylic acid, C₉H₇O₂COOH, has a pK_a = 3.480. A solution was prepared by dissolving 6.20 g of acetylsalicylic acid in 1.000 L of solution.   mw = 180.16 g/mol.

A. Calculate the pH using the quadratic and the approximate approaches.

Approximate method yields pH = 2.47

Quadratic methods yields     pH = 2.49

B   Calculate the percent error introduced in the hydrogen ion concentration if the approximate solution was chosen

% error in terms of hydrogen ion concentration is 4.98%

I know the answer for B, I just don't know how to do it :(

Answer 1

pKa = 3.480

-logKa = 3.480

Ka = 3.3 x 10^-4

Moles of Salicylic acid = given wt./molar mass

                                         = 6.20/180.16 = 0.0344 moles

Molarity of Salicylic acid = moles of salicylic acid/litres of solution

                                = 0.0344/1 = 0.0344 M

        C9H7O2COOH + H2O C9H7O2COO- + H3O+

I .....0.0344 .................................0.................0

C .....-x ....................................+x...................+.x

E ...0.0344-x........................x........................x

Approximate method

Ka = [C9H7O2COO-][H3O+]/ [C9H7O2COOH]

3.3 x 10^-4 = x²/0.0344 -x

3.3 x 10^-4 = x²/0.0344 (0.00344-x is approx. = 0.00344)

x = (3.3 x 10^-4 x 0.0344)^1/2 = 0.00337

[H+] = 0.00337

pH = -log(0.00337) = 2.47

Quadratic method

Ka = [C9H7O2COO-][H3O+]/ [C9H7O2COOH]

3.3 x 10^-4 = x²/0.0344 -x

3.3 x 10^-4 (0.0344 -x ) = x²

0.000011352 - 0.00033x = x²

x²+0.00033x -0.000011352 = 0

x = 0.00321

[H+] = 0.00321

pH = -log(0.00321) = 2.49

B ) Percent error in hydrogen ion concentration = [H+]_{approximate method} -[H+]_{quadratic method}/[H+]_{approximate method}

                                                                   = 0.00337 -0.00321/0.00337 = 4.75 %