Sulfurous acid H_2SO_3, is a diprotic acid, with K_a1 = 1.39 * 10^-2 and K_a2 =

Question

Sulfurous acid H_2SO_3, is a diprotic acid, with K_a1 = 1.39 * 10^-2 and K_a2 = 6.73 * 10^-8 Determine [H^+] for a 0.450 M sulfurous acid solution, taking into account only the first ionization. Determine [H^+] for the same sulfurous acid solution, taking into account both ionization. What is the concentration of [SO_3^2-] in the solution? What is the pH of the solution? Suppose you have 25.00 mL of a 0.240 M pyruvic acid (K_a = 3.16 Times 10^-3) if you titrate this solution with 0.100 m NaOH, at what volume of base added will you reach the equivalence point? What will be the pH of the solution at the equivalence point? What is the half-way point for the titration? What is the pH of the solution at the half-way point?

Explanation / Answer

1)

a) H2SO3 <--------> HSO3- + H+

0.45 ___________0_______0

0.45-x___________x________x

Ka1=[HSO3-][H+]/[H2SO3]= x2/0.45-x = 1.39x10-2

x= 0.072M=[H+]

b) Now the initial concentrations of the compound are the ones that we achieve at the first equilibrium:

HSO3- <---------> H+ + SO3-2

0.072M_______0.072M____0

0.072-x_______0.072+x____x

Ka2= x(0.072+x)/(0.072-x)= 6.73x10-8

The value of Ka2 is really small so we can assume that the HSO3- will dissociate a little bit and x is almost 0.

Ka2= 0.072x/0.072= x= 6.73x10-8= [H+]

final [H+]=  6.73x10-8 + 0.072 ---> this is almost 0.072M

c) [SO3-2]= 6.73x10-8M

d) pH= -log[H+]= -log(6.73x10-8 + 0.072)= 1.14

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.