Question

When the pH of 0.10 m HClO2(aq) was measured, it was found to be 1.2. a. What are the values of Ka and pKa of chlorous acid? b. The pH of a 0.10 m propylamine, C3H7NH2, aqueous solution was measured as 11.86. c. What are the values of Kb and pKb of propylamine?

Answer 1

Answer:

Explanation:

To know this, we need to write the chemical equation that is taking place here in both reactions.

a) A 0.10 M  of HClO2 solution with pH = 1.2

Let's get first with the pH the concentration of the H3O+ in equilibrium:

[H₃O⁺] = 10^(-pH)

[H₃O⁺] = 10^(-1.2) = 0.063 M

Now, let's write the equilibrium reaction for this acid:

HClO₂ + H₂O <-------> H₃O⁺ + ClO₂⁻

Now, an ICE chart for this reaction:

     HClO₂ + H₂O <-------> H₃O⁺ + ClO₂⁻

I:      0.1                                0          0

C:      -x                               +x         +x

E:   0.1-x                               x          x

With this, we can write an expression for Ka which is:

Ka = [H₃O⁺] [ClO₂⁻] / [HClO₂] = x² / 0.1 - x

However, the value of "x" we already calculate it above and it's 0.063 M, so, all we have to do is replace this value in the expression, and solve for Ka:

Ka =  (0.063)² / 0.1 - 0.063

Ka = 0.1072

Then the pKa is:

pKa = -logKa

pKa = -log(0.1072)

pKa = 0.9695

With this, we use the same procedure for part b) but instead of use pH and H₃O⁺, we will use OH⁻ and pOH.

b) 0.1 M of a C₃H₇NH₂ solution with pH = 11.86

Like in part a) we calculate the concentration of [OH⁻] instead because this is a base so, to get that, we use pOH:

14 = pH + pOH

pOH = 14 - pH

pOH = 14 - 11.86 = 2.14

[OH⁻] = 10^(-pOH)

[OH⁻] = 10^(-2.14) = 7.24x10⁻³ M

Now, we use an ICE chart again and do the same procedure as part a) so:

C₃H₇NH₂ + H₂O <-------> C₃H₇NH₃⁺ + OH⁻

I:      0.1                                0          0

C:      -x                               +x         +x

E:   0.1-x                               x          x

Kb = (7.24x10⁻³)² / (0.1 - 7.24x10⁻³)

Kb = 5.65x10⁻⁴

pKb = -log(5.65x10⁻⁴)

pKb = 3.25