Question

A 0.1276 g sample of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0633 M NaOH solution. The volume of base required to bring the solution to the equivalence point was 18.4 mL. (a) Calculate the molar mass of the acid. (b) After 10.0 mL of base had been added during the titration, the pH was determined to be 5.87. What is the Ka of the unknown acid? (10 points)

Answer 1

Ans. Step a: Calculating molar mass of acid

A weak monoprotic acid, HA reacts completely with NaOH as follow-

            HA + NaOH ------> NaA + H2O

Stoichiometry: 1 mol NaOH neutralizes 1 mol HA.

Moles of HA = Moles of NaOH consumed to reach equivalence point

                                    = Molarity of NaOH x Volume of NaOH consumed in liters

                                    = 0.0633 M x 0.0184 L                                 ; [1 L = 10³ mL]

                                    = (0.0633 mol/ L) x 0.0184 L                       ; [1 M = 1 mol/L]

                                    = 0.0011647 mol

Therefore moles of acid in 0.1276 g sample = 0.0011647 mol

Molar mass of HA = Mass of HA / Moles of HA in that mass

                                    = 0.1276 g / 0.0011647 mol

                                    = 109.55 g/mol

Step b. Moles of HA neutralized by 10 mL base = Moles of NaOH in 10 mL

                                                = Molarity of NaOH x Volume of NaOH consumed in liters

                                                = 0.0633 M x 0.010 L

                                                = 0.000633 mol

So, after addition of 10 mL NaOH, 0.000633 mol of acid would be converted into its corresponding conjugate base (A^-).

Remaining moles of HA = Initial moles of HA – moles of acid neutralized by 10 mL NaOH

                                                = 0.0011647 mol – 0.000633 mol

                                                = 0.0005317

Now,

            Total volume of titration mixture = 25.0 mL original acid solution + 10.0 mL NaOH

                                                                        = 35.0 mL = 0.035 L

            [A^-] = Moles of A^- / Volume of solution in liters

                        = 0.000633 mol / 0.035 L

                        = 0.018086 M

            [HA] = remaining moles of HA / Volume of solution in liters

                        = 0.0005317 mol / 0.035 L

                        = 0.015192 M

Now, using Henderson- Hasselbalch equation –

            pH = pKa + log ([A^-] / [AH])                    - equation 1

            where, A^- = conjugate base (sodium formate)     ; AH = Acid

Putting the values in equation 1-

            5.87 = pKa + log ([0.018086] / [0.015192]) = pKa + log 1.19048

            Or, 5.87 = pKa + 0.076

            Or, pKa = 5.87 – 0.076 = 5.79

Thus, pKa of acid = 5.79

And, using the formula:       pKa = -log Ka

            Or, 5.79 = - log Ka

            Or, Ka = antilog (- 5.79) = 1.62 x 10^-6

Thus, Ka of acid = 1.62 x 10^-6