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Question: The concentration of \[{K^ + }\] ions in the interior and exterior of a Nerve cell are \(500\) \[mM\...

The concentration of K+{K^ + } ions in the interior and exterior of a Nerve cell are 500500 mMmM and 2525 mMmM respectively. The electrical potential (in volt) that exists across the membrane is: [log102=0.3,2.303 RT/F=0.06][lo{g_{10}}2 = 0.3,2.303{\text{ }}RT/F = 0.06] (Given answer be multiplying with 10001000)

Explanation

Solution

We need to remember that the electrical potential is decided by concentration gradients of charged particles across the membrane and by membrane permeability to every sort of ion. In a resulting neuron, there are concentration gradients over the membrane for sodium ion and potassium ion. Ions move downward their gradients via channels, resulting in a separation of charge that makes the electric potential.

Complete step by step answer:
As we know that the neuron is the basic cell unit of the nervous system. Neurons are available in a variance of shapes and sizes, there are over a billion within the body.
The resting neuron features a voltage over its membrane called an electric potential, or just the resting membrane potential. Note that the charge is also because the concentration gradients influence this distribution.
The Nernst equation, at the concentration and charge forces working on a specific ion to ascertain if that ion is in passive equilibrium.
Given, Equilibrium potential for an ion (E)\left( E \right) is adequate to the universal gas constant times absolutely the temperature divided by the charge on that ion by Faraday’s constant, this obtained quantity is multiplied by the log of the concentration ration inside and exterior of the membrane is:
Given, the concentration of K+{K^ + } ion in interior of a Nerve cell is 500500 mMmM
Exterior of a nerve cell is 2525 mMmM
[log102=0.3,2.303 RT/F=0.06][lo{g_{10}}2 = 0.3,2.303{\text{ }}RT/F = 0.06]
E=0.061log[K+]exterior[K+]interiorE = \dfrac{{0.06}}{1}\log \dfrac{{{{[{K^ + }]}_{exterior}}}}{{{{[{K^ + }]}_{interior}}}}
Now we can substitute the given values we get,
E=0.061log25500E = \dfrac{{0.06}}{1}\log \dfrac{{25}}{{500}}
On simplification we get,
E=0.0780E = - 0.0780
Given answer be multiplying with 10001000
So,
(0.0780)×1000=78( - 0.0780) \times 1000 = - 78
The electrical potential (in volt) that exists across the membrane is 78 - 78.

Note:
We need to remember that the equilibrium potential for potassium is negative in some species, this suggests that the inside of the cell would need to be 70 - 70 to 75 - 75 , so as to stop the external movement of potassium given these concentration gradients. The resting membrane potential is close but not just like the K+{K^ + } ion equilibrium potential in neurons. Neuron resting membrane potentials in under physiological conditions are somewhat less negative than the K+{K^ + } equilibrium potential.