Question

Question #20ded

Answer 1

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

`DeltaPsi_(K^(+)) = (RT)/(zF) * ln(([K^(+)]_"out")/([K^(+)]_text(in)))`, where

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

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

Replace the natural log with the base 10 log to get

`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^@"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"`.

`"+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(([K^(+)]_"out")/("110 mmol L"^(-1))) = ("+52.0"cancel("mV"))/(60.2cancel("mV")) = 0.8667`

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

`[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)`, the answer will be

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