Law of Electroneutrality question Please show your work, thank you! Law of Elect
ID: 1023220 • Letter: L
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Law of Electroneutrality question Please show your work, thank you!
Law of Electroneutrality Inside the cell [Na^+] = 12 mM [K^+] = 150 mM [Ca^2+] = 0.01 mM [CI^-] = 5 mM [HCO_3^-] = 12 mM [Protein^-] = ??????? Outside the cell [Na^+] = 145 mM [K^+] = 4 mM [Ca^2+] = 2 mM [CI^-] = 105 mM [HCO_3^-] = 25 mM [Protein^-] = ??????? Solve for what [Protein"] is the teacher said that if you add up all the molar concentrations of the cations and anions you can see they don't equal zero but they must because of the protein concentrations.Explanation / Answer
Inside cell:
Total positive charge = [Na+] + [K+] + [Ca+2] = 12 mM+150mM+ (2x0.01)mM
= 162.02 mM
Total Negative charge = [Cl-]+[HCO3-1] + [Protein-] = 5mM + 12mM +[Protein-]
= 17mM +[Protein-]
In a cell, total positive charge = total negative charge
Thus, 162.02 mM = 17 mM +[Protein-]
Hence, [Protein-] = 145.02 mM
Outside cell:
Total positive charge = [Na+] + [K+] + [Ca+2] = 145 mM+4mM+ (2x2)mM
= 153 mM
Total Negative charge = [Cl-]+[HCO3-1] + [Protein-] = 105mM + 25mM +[Protein-]
= 130mM +[Protein-]
In a cell, total positive charge = total negative charge
Thus, 153 mM = 130 mM +[Protein-]
Hence, [Protein-] = 23 mM
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