<u>Given:</u>
Concentration of HNO3 = 7.50 M
% dissociation of HNO3 = 33%
<u>To determine:</u>
The Ka of HNO3
<u>Explanation:</u>
Based on the given data
[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M
The dissociation equilibrium is-
HNO3 ↔ H+ + NO3-
I 7.50 0 0
C -2.48 +2.48 +2.48
E 5.02 2.48 2.48
Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23
Ans: Ka for HNO3 = 1.23
Answer:
okay.. Questions?????????
Answer:
pOH = 4.8
pH = 9.2
Explanation:
Given data:
Hydrogen ion concentration = 6.3×10⁻¹⁰M
pH of solution = ?
pOH of solution = ?
Solution:
Formula:
pH = -log [H⁺]
[H⁺] = Hydrogen ion concentration
We will put the values in formula to calculate the pH.
pH = -log [6.3×10⁻¹⁰]
pH = 9.2
To calculate the pOH:
pH + pOH = 14
We will rearrange this equation.
pOH = 14 - pH
now we will put the values of pH.
pOH = 14 - 9.2
pOH = 4.8
Answer:
[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52
Explanation:
Kb of the reaction:
NH3 + H2O(l) ⇄ NH4+ + OH-
Is:
Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]
<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>
<em>[NH₄⁺] = [OH⁻] = X</em>
<em>And as </em>[NH₃] = 0.619M
1.8x10⁻⁵ = [X] [X] / [0.619M]
1.11x10⁻⁵ = X²
3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]
<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />
% ionization:
[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%
pH:
As pOH = -log [OH-]
pOH = 2.48
pH = 14 - pOH
<h3>pH = 11.52</h3>
I'm not sure how many sign fig's you are required to have.
However I think the final answer would be 0.05 Moles, because of the .5g, that is considered 1 sign fig.
