<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
The answer is option C.
That is it is a heterogeneous mixture.
Heterogeneous mixture have the following properties:
1. Different components could be observed in the substance.
2. Different samples of the substance appeared to have different proportions of the components.
3.The components could be easily separated using filters and sorting.
Answer:
9 moles of ions
Explanation:
Our compound is: CaCl₂(s)
We dissociate it:
CaCl₂(aq) → Ca²⁺ (aq) + 2Cl⁻(aq)
Per 1 mol of chloride, we have 1 mol of calcium cation and 2moles of chlorides, so in total we have 3 moles of ions.
Therefore in 3 moles of chloride, we would have 9 moles of ions (3 . 3)
Answer:
The pH of the buffer solution = 8.05
Explanation:
Using the Henderson - Hasselbalch equation;
pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]
where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21
Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)
[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M
[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M
Therefore,
pH = 7.21 + log (0.663 / 0.096)
pH = 7.21 + 0.84
pH = 8.05
