I need help, I don't know how to approach this question
The free energy generated by ion movement in the cell is used by the cell to do work. Compute the minimum number of millimoles of sodium that would be needed to pump one millimole of calcium ions if the following concentrations were present in the cell. Assume the temperature to be 370C and recall that R=0.008314 kJ/mol-K.
Na^+ (aq, 11mM) ----> Na^+ (aq,9.25 mM)
Ca^2+ (aq, 5mM) ----> Ca^2+ (aq, 475mM)
I understand the question, vaguely, but can you clue me in on the concentrations at the end of the post. You have two mM for both. Which is right for each?
The two chemical equations with the concentration amounts are as listed on the problem. That's what I was trying to figure out normally I can solve if only given one set of concentrations but I don't have a clue as to what to do now that I am given a mother set
I don't understand the problem either.
To compute the minimum number of millimoles of sodium needed to pump one millimole of calcium ions, we need to understand the concept of the Nernst equation and use it to calculate the electrochemical potential difference between the two ions.
The Nernst equation is used to calculate the equilibrium potential (in volts) for an ion across a membrane. In this case, we are interested in the sodium-potassium pump, which pumps sodium ions into the cell and potassium ions out of the cell. The equation is as follows:
E = (RT/zF) * ln([ion]out / [ion]in)
Where:
- E is the equilibrium potential in volts
- R is the ideal gas constant (0.008314 kJ/mol-K in this case)
- T is the temperature in Kelvin (370C = 643K in this case)
- z is the valence of the ion (+1 for sodium and +2 for calcium)
- F is Faraday's constant (96,485 C/mol)
- [ion]out and [ion]in are the ion concentrations outside and inside the cell, respectively.
Let's calculate the equilibrium potential difference for both sodium and calcium ions using the given concentrations.
For sodium ions:
E(Na) = (RT/zF) * ln([Na+]out / [Na+]in)
E(Na) = (0.008314 * 643 / 1 * 96,485) * ln(9.25 / 11)
E(Na) = -0.0252 V
For calcium ions:
E(Ca) = (RT/zF) * ln([Ca^2+]out / [Ca^2+]in)
E(Ca) = (0.008314 * 643 / 2 * 96,485) * ln(475 / 5)
E(Ca) = 0.215 V
To pump one millimole of calcium ions, we need to overcome an electrochemical potential difference of 0.215 V.
Now, to compute the minimum number of millimoles of sodium needed, we can use the equation:
ΔG = -nFΔE
Where:
- ΔG is the change in Gibbs free energy (in kJ/mol)
- n is the number of moles of electrons transferred per mole of ions (n = 2 for calcium ions)
- F is Faraday's constant (96,485 C/mol)
- ΔE is the electrochemical potential difference (0.215 V for calcium ions)
Let's calculate the value for sodium ions:
ΔG(Na) = -nFΔE
ΔG(Na) = -(2 * 96,485) * 0.215
ΔG(Na) = -41.506 kJ/mol
Now, let's find out the minimum number of millimoles of sodium ions needed:
-1 millimole of calcium ions has an associated ΔG of -41.506 kJ/mol
-1 millimole of sodium ions will have the same ΔG, as we use the same energy source
Therefore, we will need an equal number of millimoles of sodium ions, which is 1 millimole.
So, the minimum number of millimoles of sodium needed to pump one millimole of calcium ions is 1 millimole.