Question

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

Answer 1

Answer:

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:

$K_a=\frac {[H^+][B^-]} {[HB]}$

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:

$K_a=\frac {x \times x} {(C-x)}$

x² + Kₐx - CKₐ = 0

$x = \frac{-K_a + \sqrt{K_a^2 + 4CK_a}} {2}$

$x = \frac{-2.3 \times 10^{-2} + \sqrt{{(2.3 \times 10^{-2}})^2 + 4(0.135)( 2.3 \times 10^{-2})}} {2}$

So the dissociated concentration is:

x= 0.0454

The pH of the solution is:

pH= - log[0.0454]= 1.34