Question

Gastric juice is made up of substances secreted from parietal cells, chief cells, and mucous-secreting cells. The cells secrete HCI proteolytic enzyme zymogens, mucin, and intrinsic factor. The pH of gastric juice is acidic, between 1-3. If the pH of gastric juice is 2.3. what is the amount of energy (AG) required for the transport of hydrogen ions from a cell (internal pH of 7.4) into the stomach lumen? Assume that the potential difference across the membrane separating the cell and the interior of the stomach is - 70.0 mV (inside of the cell is negative relative to the lumen of the stomach). Assume that the temperature is 37 °C. The Faraday constant is 96.5 kJ/(V-mol) and the gas constant is Express your answer in kilojoules per mole. AG = kJ/mol

Answer 1

Given The pts of gastric juice is acidie, between 1-3. I the pH of gastric juice js 2.3. Internal pH of 7.4 The interior of the stomach ig - 70.0 mp The tempesatore is 37°c The paraday costant is 96.5 kJ/[rmal) Gastric juice ph = 2:3 Internal.pH of the cell = 74 :: Concentration of it inside : 10174 1 = 3.96 x 108m i concentration of #* ingide - 3.9831000M concentration of #* out side = 10-23 =5.01x1063M :-) concentration of #t out side =5.01x 162m

NOW, keq = [++ ]out side [+t] inside 3.988100m 5:01x10 3M = 7.948666 ::( keq = 7.94x106 AG = -htln keq 2-48-314 416-3 kolmalk) (310) In (7.94%104) = - 18:314 816-3 kolmol k) (310) (-11.743) AG, = 30.26 kJ/mol AG12= -FEO 5-1X/96.5 kulmol) (- 70 x y ) = 6.455 kolmod TAG2 = 6.755 kJ Imol Amout of Energy (AG) required for the transport ions form a cell (internal pH of 7.4)

into the stomach lumen = 30.26 kJ/mol + 6.75 kJ/mol = 37.01 kJ/mol. . [AG= 37.01 kJ/mol