# Question 20ded

Apr 27, 2015

Once again, this looks like a plug and play for the Nernst equation

$\Delta {\Psi}_{{K}^{+}} = \frac{R T}{z F} \cdot \ln \left(\frac{{\left[{K}^{+}\right]}_{\text{out}}}{{\left[{K}^{+}\right]}_{\textrm{\in}}}\right)$, where

$R$ - the universal gas constant;
$z$ - the valence of the ion - in your case, this will be +1;
$F$ - Faraday's constant;
${\left[{K}^{+}\right]}_{\text{out}}$ - the concentration of the potassium cations outside the cell (in the extracellular liquid);

${\left[{K}^{+}\right]}_{\text{in}}$ - the concentration of the potassium cations inside the cell (intracellular liquid).

Replace the natural log with the base 10 log to get

$\Delta {\Psi}_{{K}^{+}} = {\underbrace{2.303 \cdot \frac{R T}{F}}}_{\frac{\textcolor{b l u e}{\text{NOT 61.55 mV")) * 1/z * log(([K^(+)]_"out}}}{{\left[{K}^{+}\right]}_{\textrm{\in}}}}$

Since you're not working at ${37}^{\circ} \text{C}$, you can't use the 61.55 mV value for that product of constants. However, that will not be a problem since you've got all the values you need to solve for ${\left[{K}^{+}\right]}_{\text{out}}$.

"+52.0 mV" = 2.303 * (8.3145cancel("J")/(cancel("mol") * cancel("K")) * (273.15 + 30)cancel("K"))/(96,485cancel("J")/("V" * cancel("mol"))) * log(([K^(+)]_"out")/("110 mmol L"^(-1)))

$\log \left(\left({\left[{K}^{+}\right]}_{\text{out")/("110 mmol L"^(-1))) = ("+52.0"cancel("mV"))/(60.2cancel("mV}}\right)\right) = 0.8667$

([K^(+)]_"out")/"110 mmol L"^(-1) = 10^(0.8667) = 7.357

${\left[{K}^{+}\right]}_{\text{out" = 7.357 * "110 mmol L"^(-1) = "809.3 mmol L}}^{- 1}$

Rounded to two sig figs, the number of sig figs given for ${\text{110 mmol L}}^{- 1}$, the answer will be

[K^(+)]_"out" = color(green)("810 mmol L"^(-1))#