<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
C. Weak acid is the correct answer
Answer:
The law of constant proportions states that chemical compounds are made up of elements that are present in a fixed ratio by mass. This implies that any pure sample of a compound, no matter the source, will always consist of the same elements that are present in the same ratio by mass.
Answer:
2.94
Explanation:
There is some info missing. I think this is the original question.
<em>A solution is prepared at 25 °C that is initially 0.38 M in chloroacetic acid (HCH₂ClCO₂), a weak acid with Ka= 1.3 x 10⁻³, and 0.44 M in sodium chloroacetate (NaCH₂CICO₂). Calculate the pH of the solution. Round your answer to 2 decimal places.</em>
<em />
We have a buffer system formed by a weak acid (HCH₂ClCO₂) and its conjugate base (CH₂CICO₂⁻ coming from NaCH₂CICO₂). We can calculate the pH using the Henderson-Hasselbalch equation.
pH = pKa + log [CH₂CICO₂⁻]/[HCH₂ClCO₂]
pH = -log 1.3 x 10⁻³ + log (0.44 M/0.38 M)
pH = 2.94