Question

(a) Many biochemical reactions that occur in cells require relatively high concentrations of potassium ion $\mathbf{\left(}{\mathbf{\text{K}}}^{\mathbf{\text{+}}}\mathbf{\right)}$. The concentration of ${\mathbf{\text{K}}}^{\mathbf{\text{+}}}$in muscle cells is about $\mathbf{\text{0.15M}}$. The concentration of ${\mathbf{\text{K}}}^{\mathbf{\text{+}}}$in blood plasma is about $\mathbf{\text{0.0050M}}$. The high internal concentration in cells is maintained by pumping ${\mathbf{\text{K}}}^{\mathbf{\text{+}}}$from the plasma. How much work must be done to transport role="math" localid="1663854169668" $\mathbf{\text{1.0mole}}$of ${\mathbf{\text{K}}}^{\mathbf{\text{+}}}$from the blood to the inside of a muscle cell at${\mathbf{\text{37}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$ (normal body temperature)?

(b) When $\mathbf{\text{1.0mole}}$of ${\mathbf{\text{K}}}^{\mathbf{\text{+}}}$is transferred from blood to the cells, do any other ions have to be transported? Why or why not?

(c) Cells use the hydrolysis of adenosine triphosphate, abbreviated $\mathbf{\text{ATP}}$, as a source of energy. Symbolically, this reaction can be represented as

$\mathbf{\text{ATP}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}{\mathbf{\text{+H}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}\mathbf{\left(}\mathbf{I}\mathbf{\right)}\mathbf{\to }\mathbf{\text{ADP}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}{\mathbf{\text{+H}}}_{\mathbf{\text{2}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}^{\mathbf{-}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}$where $\mathbf{\text{ADP}}$represents adenosine diphosphate. For this reaction at ${\mathbf{\text{37}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$, role="math" localid="1663854582755" $\mathit{K}\mathbf{=}\mathbf{1}\mathbf{.7}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}$. How many moles of $\mathbf{\text{ATP}}$must be hydrolysed to provide the energy for the transport of $\mathbf{\text{1.0mole}}$of ${\mathbf{\text{K}}}^{\mathbf{\text{+}}}$? Assume standard conditions for the$\mathbf{\text{ATP}}$ hydrolysis reaction.

Short answer

(a) The work needed to be done to transport a mole of a potassium ion from the blood into a muscle cell is$-\text{8.77}\frac{\text{kJ}}{\text{mol}}$. The negative value means that there is a loss in energy.

(b) Sodium cations will be transferred in the opposite direction.

(c) $\text{0.28mol}$of$\mathrm{ATP}$ is needed for the transport of ${\text{1.0molK}}^{+}$mol ions.

Step-by-step solution

Concept Introduction:

Gibbs free energy, also known as Gibb’s function, Gibb’s energy, or free enthalpy, is a term used to measure the greatest amount of work done in a thermodynamic system while temperature and pressure remain constant.

(a) Amount of work done:

The concentration of ${\text{K}}^{\text{+}}$is $\text{0.15M}$in muscle cells, and $\text{0.0050M}$in blood plasma

Transport at normal body temperature is ${\text{37}}^{\text{o}}\text{C}$.

Calculate work done in terms of free energy ${\text{ΔG}}^{\text{o}}$.

$\begin{array}{rcl}w& =& \Delta G\\ & =& -RT\ln \left(K\right)\end{array}$

Here, the gas constant is,

$R=8.3145\raisebox{1ex}{$\text{J}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}\cdot \text{K}$}\right.$

The temperature is,

$\begin{array}{rcl}T& =& 37°\text{C}\\ & =& 37°\text{C}+273.15\\ & =& 310.15\text{K}\end{array}$

But,

$\text{K}=\frac{{\left[{\text{K}}^{\text{+}}\right]}_{\text{musclecells}}}{{\left[{\text{K}}^{\text{+}}\right]}_{\text{bloodplasma}}}$

Therefore,

$\begin{array}{rcl}\text{w}& =& -\left(\text{8.3145}\frac{\text{J}}{\text{mol}\cdot \text{K}}\right)\times \left(\text{310.15K}\right)\times \text{ln}\left(\frac{{\text{0.15MK}}^{\text{+}}}{{\text{0.0050MK}}^{\text{+}}}\right)\\ & =& -\text{8770}\frac{\text{J}}{\text{mol}}\\ & =& -8.\text{77}\frac{\text{kJ}}{\text{mol}}\end{array}$

Therefore, the value for work is $-\text{8.77}\frac{\text{kJ}}{\text{mol}}$.

(b) Ions need to be transported or not:

To maintain charge equilibrium, we have to add anions to compensate for a positive charge or remove cations to reduce the positive charge. In the human body, there is a balance between potassium and sodium ions and usually their direction is reversed regarding cell inflow and outflow.

Therefore, sodium cations will be transported.

(c) Moles of  that must be hydrolysed:

The information provided is,

$\begin{array}{rcl}T& =& \text{273K}+\text{37K}\\ & =& \text{310K}\end{array}$

$w=\text{8.8kJ}$

Calculate Gibbs free energy using the equation below.

$\begin{array}{rcl}{\text{ΔG}}^{\text{o}}& =& RT\text{ln}\left(K\right)\\ & =& \text{8.314}\frac{\text{J}}{\text{mol}\cdot \text{K}}\times \text{310K}\times \text{ln}(\text{1.7}\times {\text{10}}^{\text{5}})\\ & =& \text{31.0}\frac{\text{kJ}}{\text{mol}}\end{array}$

Now, divide the work calculated from part a of the exercise with Gibbs free energy to get the information on how many moles of $\text{ATP}$should be hydrolysed.

$\begin{array}{rcl}n& =& \frac{\text{8.8kJ}}{\text{31.0}{\displaystyle \frac{\text{kJ}}{\text{mol}}}}\\ & =& \text{0.28mol}\end{array}$

Hence, the amount of moles to be hydrolysed is $\text{0.28mol}$.