Question

Gastric juice (pH 1.5) is produced by pumping HCl from blood plasma (pH 7.4) into the stomach. Calculate the amount of free energy required to concentrate the H in 1 liter of gastric juice at 37 degree of centigrade. Under cellular conditions, how many moles of ATP must be hydrolyzed to provide this amount of free energy

Answer 1

Given question is incomplete. The complete question is as follows.

Gastric juice (pH 1.5) is produced by pumping HCl from blood plasma (pH 7.4) into the stomach. Calculate the amount of free energy required to concentrate the H in 1 liter of gastric juice at 37 degree of centigrade. Under cellular conditions, how many moles of ATP must be hydrolyzed to provide this amount of free energy? the free energy change for ATP hydrolysis under cellular conditions is about -58 kJ/mol. Ignore the effects of the transmembrane electrical potential.

Explanation:

The given data is as follows.

       Gastric juice pH = 1.5,        blood plasma pH = 7.4

    Temperature = $37^{o}C$ = (37 + 273) K

                           = 310 K

Now, relation between pH and concentration of hydrogen ions is as follows.

           pH = $-log [H^{+}]$

At pH = 1.5, we will calculate the $[H^{+}]$ as follows.

            $[H^{+}] = 10^{-pH}$

                        = $10^{-1.5}$

                        = $3.16 \times 10^{-2}$ M  ($C_{2}$)

At pH = 7.4, we will calculate the $[H^{+}]$ as follows.

            $[H^{+}] = 10^{-pH}$

                        = $10^{-7.4}$

                        = $3.98 \times 10^{-8}$ M   ($C_{1}$)

Also,

        $\Delta G_{f} = RT ln (\frac{C_{2}}{C_{1}})$

                      = $8.314 \times 310 \times ln (\frac{3.16 \times 10^{-2}}{3.98 \times 10^{-8}})$

                      = 35 kJ/mol

So, the amount of ATP necessary to provide 35 kJ is as follows.

           $\frac{35 kJ}{58 kJ/mol}$

             = 0.6 mol

Therefore, we can conclude that 0.6 moles of ATP must be hydrolyzed to provide this amount of free energy.