If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the answer to one decimal place. pKa = 4.87

pH = 4.543

Explanation:

CH3CH2COOH + H2O ↔ CH3CH2COO - + H3O + pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ C CH3CH2COOH = ((0.200 L) (0.15 M)) - ((0.300 L) (0.02 M)) / (0.3 + 0.2)

⇒ C CH3CH2COOH = 0.048 M

⇒ C NaOH = (0.300 L) (0.02 M) / (0.3 + 0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH] ... (1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M ... (2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][ ([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5) (0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O + ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543