Given, molar conductivity
\(\left(\Lambda_{m}\right)=300\, \Omega^{-1} cm ^{2} mol ^{-1}\)
Limiting molar conductivity
\(\left(\Lambda_{m}^{o}\right)=150\, \Omega^{-1} cm ^{2} mol ^{-1}\)
Concentration \((C)=0.02\, M\)
\(\therefore\) Degree of dissociation
\((\alpha)=\frac{\Lambda_{m}}{\Lambda_{m}^{o}}\)
\(\alpha=\frac{150}{300} =0.5\)
Now, by using following reaction, we can determine the value of its dissociation constant
\(K_{a}=\frac{C \alpha^{2}}{(1-\alpha)}\)
\(K_{a}=\frac{0.02 \times(0.5)^{2}}{(1-0.5)}\)
\(=\frac{0.02 \times 0.5 \times .5}{0.5}=0.01\)
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)
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm$^{-1}$ mol$^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm$^{-1}$ mol$^{-1}$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm$^{-1}$ mol$^{-1}$, its degree of dissociation is given by x $\times$ 10$^{-1}$. The value of x is ______. (Nearest integer)
Consider the following half cell reaction $ \text{Cr}_2\text{O}_7^{2-} (\text{aq}) + 6\text{e}^- + 14\text{H}^+ (\text{aq}) \longrightarrow 2\text{Cr}^{3+} (\text{aq}) + 7\text{H}_2\text{O}(1) $
The reaction was conducted with the ratio of $\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$
The pH value at which the EMF of the half cell will become zero is ____ (nearest integer value)
[Given : standard half cell reduction potential $\text{E}^\circ_{\text{Cr}_2\text{O}_7^{2-}, \text{H}^+/\text{Cr}^{3+}} = 1.33\text{V}, \quad \frac{2.303\text{RT}}{\text{F}} = 0.059\text{V}$
| Concentration of KCl solution (mol/L) | Conductivity at 298.15 K (S cm-1) | Molar Conductivity at 298.15 K (S cm2 mol-1) |
|---|---|---|
| 1.000 | 0.1113 | 111.3 |
| 0.100 | 0.0129 | 129.0 |
| 0.010 | 0.00141 | 141.0 |
An electrochemical cell is a device that is used to create electrical energy through the chemical reactions which are involved in it. The electrical energy supplied to electrochemical cells is used to smooth the chemical reactions. In the electrochemical cell, the involved devices have the ability to convert the chemical energy to electrical energy or vice-versa.