# Question d1d9f

Apr 27, 2015

As far as I can tell, all you need to do to solve this problem is use the Nernst equation

${V}_{C {a}^{2 +}} = \frac{R T}{z F} \cdot \ln \left(\frac{{\left[C {a}^{2 +}\right]}_{\text{out}}}{{\left[C {a}^{2 +}\right]}_{\textrm{\in}}}\right)$, where

$R$ - the universal gas constant;
$z$ - the valence of the ion - in your case, this will be +2;
$F$ - Faraday's constant;
${\left[C {a}^{2 +}\right]}_{\text{out}}$ - the concentration of the calcium cations outside the cell (in the extracellular liquid);

${\left[C {a}^{2 +}\right]}_{\text{in}}$ - the concentration of the calcium cations inside the cell (intracellular liquid)

The only thing to watch out for is the fact that the concentrations of the calcium cations are not given in the same unit, so you must convert one of them to the units of the other

12.0cancel("mmol")/"L"^(-1) * ("1000"mu"mol")/(1cancel("mmol")) = 12.0 * 10^(3)mu"mol L"^(-1)

The great thing about working at ${37}^{\circ} \text{C}$ is that you can use the Nernst equation like this

$V = \frac{R T}{F} \cdot \frac{1}{z} \cdot \ln \left(\frac{{\left[C {a}^{2 +}\right]}_{\text{out}}}{{\left[C {a}^{2 +}\right]}_{\textrm{\in}}}\right)$

$V = {\underbrace{2.303 \cdot \frac{R T}{F}}}_{\frac{\textcolor{b l u e}{\text{61.54 mV")) * 1/z * log(([Ca^(2+)]_"out}}}{{\left[C {a}^{2 +}\right]}_{\textrm{\in}}}}$

This means that you'll get

V_(Ca^(2+)) = "61.55 mV" * 1/2 * log((12,000cancel(mu"mol L"^(-1)))/(100cancel(mu"mol L"^(-1))))#

${V}_{C {a}^{2 +}} = \textcolor{g r e e n}{\text{+64.0 mV}}$