Question
Chemistry Question on Electrochemistry
HA is a weak acid. At 25∘C, the molar conductivity of 0.02MHA is 150Ω−1cm2mol−1. If its Λmo is 300Ω−1cm2mol−1, then equilibrium constant of HA dissociation is
0.001
0.005
0.01
0.02
0.01
Solution
Given, molar conductivity
(Λm)=300Ω−1cm2mol−1
Limiting molar conductivity
(Λmo)=150Ω−1cm2mol−1
Concentration (C)=0.02M
∴ Degree of dissociation
(α)=ΛmoΛm
α=300150=0.5
Now, by using following reaction, we can determine the value of its dissociation constant
Ka=(1−α)Cα2
Ka=(1−0.5)0.02×(0.5)2
=0.50.02×0.5×.5=0.01
The hydride of fluorine is hydrofluoric acid. Another name for it is an electron-rich hydride. Because HF possesses three lone pairs of electrons, it is electron-rich. These are within the group 17 element (halogen) category of hydride. One hydrogen atom and one fluorine atom are found in each molecule of hydrofluoric acid. When hydrogen fluoride is dissolved in water, hydrofluoric acid is produced. As a result, hydrofluoric acid is an aqueous solution of hydrogen fluoride. Fluoride from hydrogen is a very dangerous substance. If fluids are ingested, they may harm internal organs and can prove fatal if not addressed right once. If this gas is breathed, it may burn the eyes, nose, and throat and irritate those areas.
Hydrogen fluoride is formed by the two following equations:
H2 + F2 → 2HF (dark)
CaF2 + H2SO4 → 2HF + CaSO4
Hydrogen fluoride dissociates in water as follows:
HF (aq) + H2O(l) → H3O+ (aq) + F– (aq)
HF (aq) + H2O(l) ← H3O+ (aq) + F– (aq)
- Fluorescent light bulbs, aluminum, plastics, medicines, pesticides, and electrical components are all made with hydrogen fluoride. The production of refrigerants uses 60% of the hydrogen fluoride used in manufacturing.
- Using hydrogen fluoride, you can etch metal and glass.