- Darcy's Law :
- The Flow Rate Rate $Q$ is given by $Q = k * \left(\ P_B - P_A \ \right)$ , where $k$ is a constant dependent of properties of the membrane.
- Osmotic Pressure :
- $\Pi = i * M * R * T$
- $i$ = van't Hoff Factor
- $M$ = molarity
- $R$ = Gas Constant
- $T$ = temperature in Kelvin

<center><h1>Ions and Membranes</h1></center>

1. Consider a system with two compartments separated by a rigid membrane that is freely permeable to water <u>but</u> impermeable to sucrose.

<img src="https://images.34353.org/1724963483181-47808004.jpeg" alt="image-20240829163121312" style="zoom:33%;" />

- A :
- Both compartments contain pure water and pressure is applied to the piston establishing a pressure difference, $P_B > P_A$ , across the membrane.
- Describe the relation between volume flow across the membrane and the hydrostatic pressure difference , $P_B - P_A$
- The higher the pressure difference , the greater the volume of water that will flow from compartment B to A

- B :
- No force is applied to the piston and $100\ mM$ sucrose is placed in compartment B.
- Predict the direction that the meniscus (in compartment A) will move.
- The meniscus in compartment A will move downward as water flows from compartment A to B.
- This is because the sucrose creates an osmotic pressure in compartment B, drawing water across the membrane.
- Indicate the driving force for this volume flow.
- The driving force for this volume flow is the osmotic pressure difference caused by the sucrose concentration gradient.
- Water moves from an area of lower solute concentration ( compartment A ) to higher solute concentration ( compartment B ) to balance the osmotic pressure.
- Predict the compartment to add $NaCl$ ( also impermeable ) , so as to oppose this volume displacement.
- To oppose this volume displacement , $NaCl$ should be added to compartment A.
- Adding $NaCl$ to A increases the osmotic pressure in A ,

- counteracting the osmotic pressure in B
- and reducing or preventing water flow from A to B.
- Calculate the $NaCl$ concentration needed to prevent this volume displacement.
- we set the osmotic pressures equal on both sides

$$
\Pi_A = \Pi_B
$$

$$
i_{A} * M_{A} * R * T = i_{B} * M_{B} * R * T
$$

$$
i_{A} * M_{A} = i_{B} * M_{B}
$$

---

$$
2 * [\ NaCl \ ] = 1 * [\ 100\ mM \ ]
$$

$$
[\ NaCl \ ] = 50\ mM
$$
- Calculate the $MgCl_2$ concentration needed to prevent this volume displacement
- $i_{MgCl_2} = 3$

$$
3 * [\ MgCl_2 \ ] = 1 * [\ 100\ mM \ ]
$$

$$
[\ MgCl_2 \ ] = \frac{100}{3} \approx 33.33\ mM
$$
- Calculate the pressure ( and orientation ) that must be applied to the piston to prevent this volume flow.
- To prevent the volume flow, the hydrostatic pressure difference must counteract the osmotic pressure :
- $\Delta P = \Pi_B - \Pi_A = i * M * R * T$
- For $100\ mM$ sucrose in B :

$$
\Delta P = 1 * \frac{0.1\ mol}{1\ L} * \frac{0.08205736608096\ L \cdot atm}{mol \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)
$$

$$
\Delta P = 2.446540369703822\ atm * \frac{760\ mmHg}{1\ atm} = 1859.3706809749046\ mmHg
$$

- The pressure applied by the piston in B must equal this osmotic pressure to prevent flow.
- The direction of applied pressure should be from B towards A to counter the osmotic force driving water into B.

- C :
- Now consider the membrane permeable to glycerol , but less permeable to glycerol than to water. No force is applied to the piston and $100\ mM$glycerol is placed in compartment B.

- Predict the direction that the meniscus will move in the capillary of A.
- Water is drawn from A into B due to osmotic gradient
- this initially moves the meniscus down
- slowly though glycerol moves from B into A , until both A and B have equal concentrations of glycerol ( $50\ mM$ each )
- as glycerol is moving from B to A , the water is simultaneously following allong.
- The meniscus returns to the original location once the glycerol concentrations in A and B are equal

- Predict the rate of meniscus movement compare with that observed when $100\ mM$ sucrose was in compartment B.
- The rate of meniscus movement will be slower compared to when $100\ mM$ sucrose was in compartment B.
- The partial permeability of the membrane to glycerol allows some glycerol to move back into compartment A ,
- reducing the osmotic pressure difference ,
- and thus slowing the water flow.

- Calculate the pressure ( and orientation ) that must be applied to the piston to prevent this volume flow.
- The pressure needed will be less than in the sucrose case because the effective osmotic pressure difference is reduced by glycerol diffusion.

- The exact pressure depends on the rate of glycerol diffusion ,

- but can be calculated similarly to the sucrose case , adjusting for the reduced osmotic pressure.
$$
\pi_{\text{eff}} = \sigma * i * C * R * T
$$

- For a reflection coefficient of lets say $\sigma = 0.8$ for glycerol

$$
\pi_{\text{eff}} = 0.8 * 1 * 100 \, \text{mM} * R * T
$$

- The required pressure $\Delta P$ to prevent the water flow is then :

$$
\Delta P = 0.8 * 1 * 100 \, \text{mM} * R * T
$$

- The orientation should still be from B towards A to oppose the osmotic-driven water flow.

---

2. Consider two compartments of equal volume separated by a membrane ( shown below).

<img src="https://images.34353.org/1724961371420-748639006.jpeg" alt="Membrane A = 10 mM KCl || Membrane B = 100 mM KCl" style="zoom:33%;" />

- Consider the following issues for each, and any, permeability condition of the membrane.
- Comparing side A with side B, predict the orientation of the electrical potential difference (ePD) across the membrane during the initial moments of flow (before noticeable changes in chemical composition have occurred).
- 4 Issues :
1. The orientation of the ePD during the initial moments of flow.
2. The magnitude of the ePD during the initial moments.
3. The ePD at electrochemical equilibrium.
4. The chemical composition at electrochemical equilibrium.
- Condition A: The membrane is equally permeable to both $K^+$ and $Cl^-$
1. Orientation : Since both ions diffuse equally from high to low concentration, there is no net charge separation. Both K+K+ and Cl−Cl− move from side B to side A, maintaining electrical neutrality. Therefore, no significant ePD develops; the orientation is neutral (ePD = 0 mV).
2. Magnitude : The ePD remains at 0 mV because there's no charge imbalance.
3. ePD at equalibrium : remains at 0 mV because the distribution of charges is equal on both sides.
4. Chemical Composition: The ions reach equilibrium concentrations due to equal diffusion. Since volumes are equal, the final concentration in each compartment is the average:

$$
[K^+] = [Cl^-] = \frac{10\ mM + 100\ mM}{2} = 55\ mM
$$

- Condition B: The permeability of the membrane to $K^+$ is made greater than that to $Cl^-$
1. Orientation : K+ ions diffuse from side B to side A faster than $Cl^-$ ions. This leads to:
- Side A: Accumulation of positive charge ( $K^+$ influx).
- Side B: Relative negative charge due to loss of $K^+$ and slower $Cl^-$ movement.
2. Magnitude : We can estimate the ePD using the Nernst equation for $K^+$ :

$$
E_{K^+} = \frac{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}{1 * \frac{96,485\ Coulombs}{1\ mol}} * ln\left(\ \frac{100\ mM}{10\ mM} \ \right) = 61.537\ Volts
$$
- However, since Cl−Cl− is partially permeable, the actual ePD will be less than +61.537 mV.

3. ePD at equalibrium : A steady ePD develops, somewhere between 0 mV and +61.537 mV, depending on the relative permeabilities. The exact value requires the Goldman-Hodgkin-Katz equation.
4. Chemical Composition:
- $K^+$ concentration difference decreases due to its higher permeability
- $Cl^-$ concentrations adjust less significantly.
- The compartments do not reach equal ion concentrations.

- Condition C: The membrane is made impermeable to $Cl^-$ , but remains permeable to $K^+$
1. Orientation :$K^+$ ions move from side B to side A, but $Cl^-$ cannot follow. This results in:
- Side A: Positive charge accumulation ($K^+$ influx without $Cl^-$balance).
- Side B: Negative charge accumulation due to loss of $K^+$.
2. Magnitude : The ePD quickly approaches the Nernst potential for $K^+$ : $E_K \approx +61.537\ mV$
3. ePD at equalibrium : The membrane potential stabilizes at +61.537 mV (side A positive relative to side B).
4. Chemical Composition:
- Minimal change in ion concentrations due to the rapid development of ePD.
- ($K^+$ movement is limited by the electric potential that opposes further diffusion.
- $[K^+]_A$ remains close to 10 mM; $[K^+]_B$ remains close to 100 mM.

| Condition | Initial ePD Orientation | Initial ePD Magnitude | ePD at Equilibrium | Chemical Composition at Equilibrium |
| --------- | ----------------------- | --------------------- | ----------------------- | -------------------------------------------------------- |
| A | Neutral , ePD = 0 mV | 0 mV | 0 mV | Both sides : 55 mM KCl |
| B | Side A = positive | Less than +61.537 mV | Positive , < +61.537 mV | Unequal concentrations; Potassium partially equilibrated |
| C | Side A = positive | Approaches +61.537 mV | +61.537 mV | Concentrations remain close to initial values |

### **Understanding Electrochemical Equilibrium vs. Equal Concentrations**

**Electrochemical Equilibrium** occurs when the net movement of ions across the membrane stops because the electrical and chemical driving forces balance each other out. This doesn't necessarily mean that the concentrations of ions on both sides of the membrane are equal. Instead, it means that the **electrochemical potential** for each ion is the same on both sides, so there's no net driving force for further ion movement.

**Equal Concentrations** imply that the concentration of each ion is the same on both sides of the membrane. This is a special case that only occurs under specific conditions, such as when ions diffuse freely without any opposing electrical potential.

### Why Concentrations Don't Equalize in Condition B

- **Limited $\text{Cl}^-$ Movement:**
- Since $\text{Cl}^-$ moves slowly, it cannot keep up with $\text{K}^+$ to maintain electrical neutrality purely through diffusion.

- **Electrochemical Gradient Stops Net Ion Movement Before Equal Concentrations Are Reached:**
- The system reaches a point where the desire of $\text{K}^+$ to move down its concentration gradient is exactly opposed by the electrical potential difference.
- At this point, even though there's still a concentration difference, there's no net movement of $\text{K}^+$.

- **Final Concentrations:**
- $\left[ \text{K}^+ \right]_A$: Increases from its initial value but does not reach the concentration in Side B.
- $\left[ \text{K}^+ \right]_B$: Decreases slightly.
- $\left[ \text{Cl}^- \right]$: Changes minimally due to low permeability.

In Condition B, the system does reach equilibrium—specifically, electrochemical equilibrium. However, this equilibrium doesn't require equal ion concentrations across the membrane. Instead, it occurs when the electrical potential difference balances the concentration gradients, resulting in no net ion movement.

Key Takeaways:

- Equilibrium means no net ion movement, not necessarily equal concentrations.
- Differential permeability leads to charge separation and development of ePD.
- Electrochemical forces dictate the final state of the system.
- Ion concentrations at equilibrium are determined by the balance of electrical and chemical forces, not by achieving equal concentrations.

---

3. Consider two compartments of equal volume separated by a membrane (shown below).

<img src="https://images.34353.org/1724961485023-120849913.jpeg" alt="Membrane A = 100 mM KCl and 10 mM NaCl || Membrane B = 10 mM KCl and 100 mM NaCl" style="zoom:33%;" />

- Consider the 4 issues for each permeability condition :
1. The orientation of the ePD during the initial moments of flow.
2. The magnitude of the ePD during the initial moments.
3. The ePD at electrochemical equilibrium.
4. The chemical composition at electrochemical equilibrium.

| Condition | Initial ePD Orientation | Initial ePD Magnitude | ePD at Equilibrium | Chemical Composition at Equilibrium |
|-----------|-------------------------|-----------------------|-----------------------|--------------------------------------------------------------|
| D | A negative relative to B | Approximately -59 mV | Approximately -59 mV | Minimal changes; slight movement of K⁺ from A to B |
| E | A positive relative to B | Approximately -59 mV | Approximately -59 mV | Minimal changes; slight movement of Na⁺ from B to A |
| F | A positive relative to B | Approximately +41 mV | Approximately +41 mV | Gradual changes; Na⁺ accumulates in A, K⁺ accumulates in B |

---

4. The table below lists, for a particular cell type, the intracellular (i) and extracellular (e) concentrations for a number of solutes, X [ in mM ] , including monovalent and divalent ions.

- The electrical potential difference across the membrane ( $V_m$ ) of this cell type is $−60\ mV$ , cell interior with respect to exterior, at 29°C.
- A :
- Calculate the Nernst potential , $E_\text{solute}$ , for each solute in mV.

$$
E_{A^+} = \frac{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}{1 * \frac{96,485\ Coulombs}{1\ mol}} * ln\left(\ \frac{104\ mM}{7\ mM} \ \right) = 0.0693311688311402\ Volts
$$

$$
E_{E^+} = \frac{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}{1 * \frac{96,485\ Coulombs}{1\ mol}} * ln\left(\ \frac{8\ mM}{110\ mM} \ \right) = -0.06734147917183664\ Volts
$$

$$
E_{G^{++}} = \frac{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}{2 * \frac{96,485\ Coulombs}{1\ mol}} * ln\left(\ \frac{0.01\ mM}{1\ mM} \ \right) = -0.059159553324716355\ Volts
$$

$$
E_{J^-} = \frac{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}{-1 * \frac{96,485\ Coulombs}{1\ mol}} * ln\left(\ \frac{10\ mM}{5\ mM} \ \right) = -0.017808800080822428\ Volts
$$

$$
E_{L^-} = \frac{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}{-1 * \frac{96,485\ Coulombs}{1\ mol}} * ln\left(\ \frac{100\ mM}{10\ mM} \ \right) = -0.05915955332471636\ Volts
$$

$$
E_{M^-} = \frac{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}{-1 * \frac{96,485\ Coulombs}{1\ mol}} * ln\left(\ \frac{2\ mM}{2\ mM} \ \right) = 0\ Volts
$$

- $Q$ and $R$ are neutral

- B :
- Considering only the extracellular concentration $[\ X\ ]_e$ , predict the intracellular concentration$[\ X\ ]_i$ of each solute, assuming a passive distribution across the membrane (electrochemical equilibrium according to the measured $V_m$ )
$$
[\ X\ ]_i^{\text{predicted}} = [\ X\ ]_e * e^{\frac{z * F * V_m}{R * T}}
$$

( 104.0 ) * ( math.e**( ( 1.0 * 96485.0 * -0.06 ) / ( 8.31446261815324 * ( 25.0 + 273.15 ) ) ) )

$$
[\ A^+\ ]_i^{\text{predicted}} = 104\ mM * e^{\frac{1 * \frac{96,485\ Coulombs}{1\ mol} * -0.06\ V}{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}} = 10.065304057151284\ mM
$$

$$
[\ E^+\ ]_i^{\text{predicted}} = 8\ mM * e^{\frac{1 * \frac{96,485\ Coulombs}{1\ mol} * -0.06\ V}{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}} = 0.7742541582424064\ mM
$$

$$
[\ G^{++}\ ]_i^{\text{predicted}} = 0.01\ mM * e^{\frac{2 * \frac{96,485\ Coulombs}{1\ mol} * -0.06\ V}{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}} = 0.09366710961807145* \ mM
$$

$$
[\ J^{-}\ ]_i^{\text{predicted}} = 10.0\ mM * e^{\frac{-1 * \frac{96,485\ Coulombs}{1\ mol} * -0.06\ V}{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}} = 103.32524423453377 * \ mM
$$

$$
[\ L^{-}\ ]_i^{\text{predicted}} = 100.0\ mM * e^{\frac{-1 * \frac{96,485\ Coulombs}{1\ mol} * -0.06\ V}{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}} = 1033.2524423453376 * \ mM
$$

- For $M^-$ , $V_m = 0$ , therefore the predicted concentration equals the extracellular concentration = $2\ mM$

$$
[\ Q\ ]_i^{\text{predicted}} = 4.0\ mM * e^{\frac{0.0 * \frac{96,485\ Coulombs}{1\ mol} * -0.06\ V}{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}} = 4.0 * \ mM
$$

$$
[\ R\ ]_i^{\text{predicted}} = 1.0\ mM * e^{\frac{0.0 * \frac{96,485\ Coulombs}{1\ mol} * -0.06\ V}{\frac{8.31446261815324\ J}{mol\ \cdot\ ^{\circ}K} * \left(\ 25 ^{\circ}C + 273.15 \ \right)^{\circ}K}} = 1.0 * \ mM
$$
- C :
- Predict physiological cellular properties concerning these solutes (from this data), especially for those solutes that are not distributed in accord with electrochemical equilibrium.

| Solute X | $[\ X\ ]_e$ | $[\ X\ ]_i$ | $[\ X\ ]_i^{\text{predicted}}$ | $E_x$ | Predictions |
| -------- | ----------- | ----------- | ------------------------------ | --------------------- | ----------- |
| $A^+$ | 104 | 7 | 10.065304057151284 | 0.0693311688311402 | |
| $E^+$ | 8 | 110 | 0.7742541582424064 | -0.06734147917183664 | |
| $G^{++}$ | 0.01 | 1 | 0.09366710961807145 | -0.059159553324716355 | |
| $J^-$ | 10 | 5 | 103.32524423453377 | -0.017808800080822428 | |
| $L^-$ | 100 | 10 | 1033.2524423453376 | -0.05915955332471636 | |
| $M^-$ | 2 | 2 | 2 | 0.0 V | |
| $Q$ | 4 | 4 | 4 | 0.0 V | |
| $R$ | 1 | 3 | 1 | 0.0 V | |

<center><h2>Nernst Equation</h2></center>

@ $T=302.4\ ^{\circ}K = 29.2\ ^{\circ}C = 84.6 \ ^{\circ}F$
$$
E_j = -\left(\ \frac{R * T}{z_j * F} \ \right) * ln\left(\ \frac{C_2}{C_1} \ \right) = -\left(\ \frac{R * T}{z_j * F} \ \right) * ln\left(\ 10 \ \right) * log_{10}\left(\ \frac{C_2}{C_1} \ \right)
$$

$$
E_j = -\left(\ \frac{60\ mV}{z_j} \ \right) * log_{10}\left(\ \frac{C_2}{C_1} \ \right)
$$

---

$$
e^{x} = 10^{\frac{x}{log_{10}\left(\ e \ \right)}}
$$

$$
\text{antilog}_{10}\left(\ x \ \right) = e^{\left(\ x * log_e\left(\ 10 \ \right) \ \right)}
$$

---

$$
C_2 = C_1 * \exp\left(\ \frac{-z_j * F * E_j}{R * T} \ \right)
$$

$$
C_2 = C_1 * 10^{\frac{-z_j * E_j}{60\ mV}} = C_1 * \text{antilog}_{10}\left(\ \frac{-z_j * E_j}{60\ mV} \ \right)
$$