Question

4) In a titration of a 100.0mL 1.00M HA weak acid solution with 1.00M NaOH, what is the pH of the solution after the addition of 49.5 mL of NaOH? K_a = 1.80 x 10^-5 for HA. (3 significant figures)

**Remember to calculate the equivalence volume and think about in which region along the titration curve the volume of base falls, region 1, 2, 3, or 4**

5) In a titration of a 100.0mL 1.00M HCl strong acid solution with 1.00M NaOH, what is the pH of the solution after the addition of 30.9 mL of NaOH? (3 significant figures)

**Remember to calculate the equivalence volume and think about in which region along the titration curve the volume of base falls, region 1, 2, 3, or 4**

Answer 1

4. when we add NaOH to weak acid sodium salt of weak acid genarated along with some unreacted weak acid will be present. so, it will be a acidic buffer. we need to calculate the concentration of salt and un reacted acid

total voume of solution = 149.5 ml

salt concentration [NaA] = 49.5 ml * 1 M / 149.5 = .331 M

unreacted weak acid will be = 100- 49.5 = 50.5 ml 1 M

concentration of weak acid [HA] = 50.5 ml * 1 M / 149.5 = .337

pH is calculated using Henderson-Hasselbalch Equation,

   

now pH = - log k_a + log .331/.337

          =   - log 1.80 x 10^-5 - 7.8 x 10^-3

            ⁼ 4.732

5. unreacted HCl is ( 100 - 30.9 ) = 69.1 ml

Total volume of the solution is 100 + 30.9 = 130.9

concentration of HCl = 69.1 ml * 1 M / 130.9 = .521 M

pH = - log [H⁺]

   =0.277