Determine the pH of a 0.034 M solution of H₂SO₄.The dissociation occurs in two steps.K_a1 is extremely large;K_a2 is 1.2 $\times$ 10^-². A)12.66 B)1.47 C)1.92 D)1.38 E)2.12

Question

Quizplus · Accepted Answer

First, we need to write the two dissociation equations for HS1U1B12S1U1B0SOS1U1B14S1U1B0:

HS1U1B12S1U1B0SOS1U1B14S1U1B0 ⇌ H+ + HS1U1B1- + OHS1U1B14S1U1B0 (Ka1)

HS1U1B1- + H2O ⇌ HSO1U1B13S1U1B0 + OH-    (Ka2)

Note that Ka1 is extremely large, which implies that the first dissociation is almost completely accomplished.

So, we can treat [HS1U1B1-] as negligible and assume that all of the HS1U1B12S1U1B0SOS1U1B14S1U1B0 dissociates into H+ and OHS1U1B14S1U1B0.

Now we can write the expression for the equilibrium constant for the second dissociation reaction (Ka2):

Ka2 = [HSO1U1B13S1U1B0][OH-]/[HS1U1B1-]

We can assume that the [OH-] concentration is provided by the dissociation of water (1x10^-14/[OH-]), and we can calculate [HSO1U1B13S1U1B0] using the initial concentration and the fact that [HSO1U1B13S1U1B0]= [H+]:

[HSO1U1B13S1U1B0] = [H+] = (Ka1*[HS1U1B12S1U1B0SOS1U1B14S1U1B0])^1/2

[HSO1U1B13S1U1B0] = (1.2x10^-10*(0.034))^1/2 = 2.90x10^-7

[OH-] = 1x10^-14/[H+] = 1x10^-14/(Ka1*[HS1U1B12S1U1B0SOS1U1B14S1U1B0])^1/2 = 5.13x10^-6

Now we can calculate the pH:

pH = -log[H+] = -log(Kw/[OH-]) = -log(1x10^-14/5.13x10^-6) = 1.38

Therefore, the answer is D) 1.38.

Determine the PH of a 0 ×\times× 10-2

Determine the PH of a 0 $\times$ 10^-²