Question #20ded

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Question #20ded

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Answer 1 · 2015-04-27T23:54:01

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

$Δ Ψ_{K^{+}}^{+} = \frac{R T}{z F} \cdot ln (\frac{{[K^{+}]}_{out}^{+}}{{[K^{+}]}_{in}^{+}})$ $DeltaPsi_(K^(+)) = (RT)/(zF) * ln(([K^(+)]_"out")/([K^(+)]_text(in)))$ , where

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

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

Replace the natural log with the base 10 log to get

$Δ Ψ_{K^{+}}^{+} = 2.303 \cdot \frac{R T}{F}      NOT 61.55 mV \cdot \frac{1}{z} \cdot log (\frac{{[K^{+}]}_{out}^{+}}{{[K^{+}]}_{in}^{+}})$ $DeltaPsi_(K^(+)) = underbrace(2.303 * (RT)/F)_(color(blue)("NOT 61.55 mV")) * 1/z * log(([K^(+)]_"out")/([K^(+)]_text(in)))$

Since you're not working at $37^{\circ} C$ $37^@"C"$ , 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 ${[K^{+}]}_{out}^{+}$ $[K^(+)]_"out"$ .

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

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

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

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

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

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

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Question #20ded

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