Question

Calculate the pH of 0.242 M piperazine (C₄H₁₀N₂, a di-basic compound, "B"). Calculate the concentration of each form of piperazine in this solution. (K_a1 = 4.65 ✕ 10⁻⁶ and K_a2 = 1.86 ✕ 10⁻¹⁰.)
NOTE: Use three significant figures in all answers and use quadratic formula.

pH
[B]	M
[BH+]	M
[BH22+]	M

Answer 1

B + H₂O <----> BH⁺ + OH^-

Now, as OH^- is formed , therefore we need K_b values

Thus, K_a2 = K_b1 & K_a1 = K_b2

Now, K_a*K_b = K_w

Thus, Kb₁ = 5.376*10^-5 & K_b2 = 2.15*10^-9

Now,

K_b1 = {[BH⁺]*[OH^-]}/[B]

or, 5.376*10^-5 = x²/(0.242-x)

or, x = 0.0073 M

Thus, x = [BH⁺] = [OH^-] = 0.0073 M

& [B] = 0.242-0.0073 = 0.2347 M

Also, BH⁺ + H₂O <-----> BH₂²⁺ + OH^-

K_b2 = {[BH₂²⁺]*[OH^-]}/[BH⁺]

or, 2.15*10^-9 = (0.0073+x)*x/(0.0073-x)

or, x = 2.15*10^-9 approx.

Thus,

[B]	= 0.2347 M
[BH+]	= 0.0073 M
[BH22+] = 2.15*10-9 M