Question

Question1: A 100.0 mL solution of 0.5 M histidine in its tribasic form, A^3-, (pK_a1 = 1.70, pK_a2 = 6.02, pK_a3 = 9.08) is titrated with 1.0 M HCl titrant.

1)calculate the volume of titrant required to reach the second equivalence point,

2)calculate the pH after the addition of 50.0 mL of titrant,

3)calculate the pH after the addition of 62.0 mL of titrant,

4)calculate the pH after the addition of 75.0 mL of titrant.

Question4: A 100 mL volume of a 0.100 M weak base (pK_b = 5.00) was titrated with 1.00 M HClO₄. Find the pH at the following volumes of added acid: 0.00, 1.00, 5.00, 10.00, and 10.10 mL.

Question5: A 100.00 mL volume of 0.0400 M propionic acid (CH₃CH₂COOH; K_a = 1.34 ´ 10^-5) was titrated with 0.0837 M NaOH. Calculate the pH after the addition of: 0 V_e, 0.25 V_e, V_e, and 1.1 V_e mL of titrant where V_e is the volume of NaOH required to reach the equivalence point.

Answer 1

Q1:

(a)

millimole of A^3-= 100.0ml × 0.500M = 50.0mmol

Millimole of HCl needed till 2nd equivalence point = 50.0 ×2 = 100mmol

Volume of HCl used = 100.0mmol/ 1.0M = 100.0ml. (Answer)

(b)

after addition of 50.0ml of HCl , first equivalent point is reached .

pH of HA^2- solution = (pKa2 + pKa3)/2 = (6.02 + 9.08)/2 = 7.55 (answer)

(c)

After addition of 62.0ml of HCl , millimole of HCl added = 62.0ml × 1.0M = 62.0mmol

Millimole of HA^2- = 50 -12 = 38mmol

Millimole of H₂A^- = 12 mmol

pH = pKa2 + log(salt/acid) = 6.02 + log(38/12) = 6.52 (answer)

(d)

After addition of 75.0ml of HCl , millimole of HCl added = 75.0mmol

Millimole of HA^2- = 50-25 = 25mmol

Millimole of H₂A^- = 25mmol

pH = pKa2 + log(25/25) = 6.02 (Answer)