Find the ph of a 0.135m aqueous solution of periodic acid(hio4) for which ka= 2.3 x 10^-2

a. 1.17
b. 3.28
c. 1.34
d. 1.23
e. 1.84

The answer is C. pH= 1.34

Step-by-step explanation:

Periodic acid dissociates in water as:

HIO4 ⇒ H⁺ +  IO₄⁻

The formula for Ka is:

where:

[H⁺] = concentration of H+ ions

[B-] = concentration of conjugate base ions

[HB] = concentration of undissociated acid molecules

for a reaction HB ⇒  H⁺ + B⁻

Periodic acid dissociates one H⁺ ion for every  IO₄⁻ ion, so [H⁺] = [  IO₄⁻].

Let x represent the concentration of H⁺ that dissociates from HB, then [HB] = C - x where C is the initial concentration.

Then, by using the Kₐ formula:

x² + Kₐx - CKₐ = 0

So the dissociated concentration is:

x= 0.0454

The pH of the solution is:

pH= - log[0.0454]= 1.34

step-by-step explanation: i dont