Question

Your job is to determine the properties of a brand-new Brønsted-Lowry acid which we’ll refer to as ‘HQ’. You prepare a 0.100 M solution of this new acid HQ and measure the pH at 25°C. The pH of your solution is unique. Every student in the class may have a different pH value.

My pH value is 3.06

1. Based on your unique pH, determine the equilibrium concentration of [H3O+]E in your solution.

2. Determine the equilibrium concentrations of [A−]E and [HA]E in your solution.

3. Calculate the acid ionization constant Ka for this new acid HQ. You then titrate 50.0 mL of this solution of HQ (aq) with a 0.250 M solution of the strong base, NaOH (aq).

4. Determine the pH at the half-equivalence point of the titration.

5. Calculate the volume of NaOH that must be added to reach equivalence.

6. Calculate the pH of your solution at the equivalence point.

7. Calculate the pH of the solution after adding a total of 25.0 mL of NaOH(aq).

8. Draw an atomic-scale diagram showing the progress of the titration, highlighting the major species (other than water) at each of the key stages you calculated the pH for in the questions above.

★ At each stage, your diagram should clearly show (i) the total solution volume, (ii) the relative amount of hydroxide added, (iii) the compounds present before and after the neutralization reaction, and (iv) be labeled with any major equilibrium reactions occurring in solution.

Answer 1

HQ is an monoprotic acid

1. pH = -log[H3O+] = 3.06

equilibrium [H3O+] = 8.71 x 10^-4 M

2. equilibrium [A-] = 8.71 x 10^-4 M

equilibrium [HA] = 0.1 - 8.71 x 10^-4 = 0.09913 M

3. Ka = (8.71 x 10^-4)^2/0.09913 = 7.653 x 10^-6

titrate 25 ml of acid with 0.25 M NaOH

4. pH at half-equivalence = pKa = -logKa = -log(7.653 x 10^-6) = 5.12

5. NaOH volume to reach equivalence point = 0.1 M x 25 ml/0.25 M = 10 ml

6. pH at equivalence point

[A-] = 0.1 M x 25 ml/35 ml = 0.07143 M

A- + H2O <==> HA + OH-

let x amount reacted

Kb = 1 x 10^-14/7.653 x 10^-6 = x^2/0.07143

x = [OH-] = 9.66 x 10^-6 M

pOH = -log[OH-] = 5.015

pH = 14 - pOH = 8.985

7. pH after 25 ml NaOH added

excess [OH-] = 0.25 M x 15 ml/50 ml = 0.075 M

pOH = -log[OH-] = 1.125

pH = 14 - pOH = 12.875

initial pH of solution,

7.653 x 10^-6 = x^2/0.1

x = 8.75 x 10^-4 M

pH = -log[H+] = 3.06