Answer:
Explanation:
To know this, we need to write the chemical equation that is taking place here in both reactions.
<u>a) A 0.10 M of HClO2 solution with pH = 1.2</u>
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.
<u><em>b) 0.1 M of a C₃H₇NH₂ solution with pH = 11.86</em></u>
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
