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

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;