Kohlrausch Law || Electrochemistry || Molar Conductivity || Equivalent Conductivity
| CSIR-NET
Chemistry Untold
Foreign, I have taken one mole of electrolyte. When the number of moles of
electrolyte is fixed, the number of moles of electrolytes is also fixed. However,
the number of ions is fixed. One more logarithmic formula is: 6.022 x 10^23 + λ
mole is giving you x times a+ and y times b-. The sum of molar conductivity at
infinite dilution is x * λ0 of a+ + y * λ0 of b-. Each ion makes a definite
contribution towards the equivalent conductance of the electrolyte, regardless of
the nature of the ion it is associated with. The value of equivalent conductance at
infinite dilution for any electrolyte is the sum of the contributions of its own
constituent ions. The conductivity of ions of an electrolyte at infinite dilution
is constant and does not vary.
We speak in terms of normality. Can we have a relation between molarity and
normality? Can we have a relation between molar conductivity? The formula for molar
conductivity is: conductivity divided by concentration multiplied by 100. If you
remember this formula, concentration * molar conductivity or equivalent
conductivity formula = conductivity divided by normality multiplied by conductivity
of AlCl3 at infinite dilution.
The molar conductivity of AlCl3 at infinite dilution is: λ0 of Al3+ multiplied by q
plus λ0 of AlCl3 molar. This is molar capital M small molarity molality divided by
3 * (1/3) * λ0 of Al3+. Let us associate it. It's not a perfectly applicable
formula, we use cholera which is applicable only for strong electrolyte to predict
the molar conductivity at infinite dilution.
Can we have the molar conductivity of NaCl? Sorry, absolutely. Can we have the
molar conductivity of CH3COONa (sodium acetate)? Sorry, absolutely. Can I do λ0 of
HCl? Can I do λ0 of HCl plus λ0 of CH3COONa plus λ0 of NaCl? What is basically λ of
HCl? It's λ minus λ0 of Na+ minus λ0 of Cl-. Now if you look very carefully, can I
say λ0 of H+ plus λ0 of CH3COO- minus λ0 of any electrolyte can be calculated with
the help of cholera's law, which is applicable to strong electrolytes. With the
help of strong electrolytes, we can easily calculate the maximum molar conductivity
or molar conductivity at infinite dilution or zero concentration. These are the
values given to you: sodium sulphate, potassium sulfate, KCl. The λ value will be
λ0 (100% associated) - 100% dissociate.
Can you correlate it? What do you think α formula degree of dissociation or degree?
This will be k = α * C / (1 / α - 1). Establishment divided by α upper B divided by
α alpha alpha cancel out one minus α upon α, so one upon α minus α upon α is one.
This will be equal to 1 + λ / λ0 * C / Ka. λ0 / λ will be equal to 1 + λ * C / λ0 *
Ka.
Why did I split this? It's my variable. It starts from a cut on the y-axis and it
goes up like this. What will be the slope? The slope will be 1 / λ0. You may not
see many questions on this, but yes, there may be questions on CS in CSIR, IIT JAM,
or GATE. Obviously, music, God bless you.
| CSIR-NET
Chemistry Untold
Foreign, I have taken one mole of electrolyte. When the number of moles of
electrolyte is fixed, the number of moles of electrolytes is also fixed. However,
the number of ions is fixed. One more logarithmic formula is: 6.022 x 10^23 + λ
mole is giving you x times a+ and y times b-. The sum of molar conductivity at
infinite dilution is x * λ0 of a+ + y * λ0 of b-. Each ion makes a definite
contribution towards the equivalent conductance of the electrolyte, regardless of
the nature of the ion it is associated with. The value of equivalent conductance at
infinite dilution for any electrolyte is the sum of the contributions of its own
constituent ions. The conductivity of ions of an electrolyte at infinite dilution
is constant and does not vary.
We speak in terms of normality. Can we have a relation between molarity and
normality? Can we have a relation between molar conductivity? The formula for molar
conductivity is: conductivity divided by concentration multiplied by 100. If you
remember this formula, concentration * molar conductivity or equivalent
conductivity formula = conductivity divided by normality multiplied by conductivity
of AlCl3 at infinite dilution.
The molar conductivity of AlCl3 at infinite dilution is: λ0 of Al3+ multiplied by q
plus λ0 of AlCl3 molar. This is molar capital M small molarity molality divided by
3 * (1/3) * λ0 of Al3+. Let us associate it. It's not a perfectly applicable
formula, we use cholera which is applicable only for strong electrolyte to predict
the molar conductivity at infinite dilution.
Can we have the molar conductivity of NaCl? Sorry, absolutely. Can we have the
molar conductivity of CH3COONa (sodium acetate)? Sorry, absolutely. Can I do λ0 of
HCl? Can I do λ0 of HCl plus λ0 of CH3COONa plus λ0 of NaCl? What is basically λ of
HCl? It's λ minus λ0 of Na+ minus λ0 of Cl-. Now if you look very carefully, can I
say λ0 of H+ plus λ0 of CH3COO- minus λ0 of any electrolyte can be calculated with
the help of cholera's law, which is applicable to strong electrolytes. With the
help of strong electrolytes, we can easily calculate the maximum molar conductivity
or molar conductivity at infinite dilution or zero concentration. These are the
values given to you: sodium sulphate, potassium sulfate, KCl. The λ value will be
λ0 (100% associated) - 100% dissociate.
Can you correlate it? What do you think α formula degree of dissociation or degree?
This will be k = α * C / (1 / α - 1). Establishment divided by α upper B divided by
α alpha alpha cancel out one minus α upon α, so one upon α minus α upon α is one.
This will be equal to 1 + λ / λ0 * C / Ka. λ0 / λ will be equal to 1 + λ * C / λ0 *
Ka.
Why did I split this? It's my variable. It starts from a cut on the y-axis and it
goes up like this. What will be the slope? The slope will be 1 / λ0. You may not
see many questions on this, but yes, there may be questions on CS in CSIR, IIT JAM,
or GATE. Obviously, music, God bless you.
