The volume of an adult's stomach ranges from about 50 mL when empty to 1 when full. If the stomach volume is 400 mL and its contents have a pH of \(2,\) how many moles of \(\mathrm{H}^{+}\) does the stomach contain? Assuming that all the \(\mathrm{H}^{+}\) comes from \(\mathrm{HCl}\) , how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?

The pH value is related to the concentration of \(\mathrm{H}^{+}\) ions in a solution. The equation that links these two quantities is: $$ \mathrm{pH} = -\log_{10} [\mathrm{H}^{+}] $$ where \(\mathrm{pH}\) is the pH value and \([\mathrm{H}^{+}]\) represents the concentration of \(\mathrm{H}^{+}\) ions in moles per liter (M). We know that the pH of the stomach contents is 2. Using this information, we can calculate the concentration of \(\mathrm{H}^{+}\) ions in the stomach as follows: $$ [\mathrm{H}^{+}] = 10^{-\mathrm{pH}} = 10^{-2} = 0.01\,\text{M} $$

To find the number of moles of \(\mathrm{H}^{+}\) ions in the stomach, we can use the following equation: $$ \mathrm{moles\ of\ H}^{+} = [\mathrm{H}^{+}] \times \mathrm{volume\ of\ stomach\ contents} $$ We know that the concentration of \(\mathrm{H}^{+}\) ions is \(0.01\,\text{M}\) and the volume of the stomach contents is \(400\,\mathrm{mL}\). We need to convert the volume from milliliters to liters: $$ 400\,\mathrm{mL} = 400 \times 10^{-3}\,\mathrm{L} = 0.4\,\mathrm{L} $$ Now we can calculate the number of moles of \(\mathrm{H}^{+}\) ions in the stomach: $$ \mathrm{moles\ of\ H}^{+} = 0.01\,\text{M} \times 0.4\,\mathrm{L} = 0.004\,\text{moles} $$

To find the mass of sodium hydrogen carbonate needed to neutralize the stomach acid, we must first know the balanced chemical equation for the reaction between hydrochloric acid and sodium hydrogen carbonate. This equation is: $$ \mathrm{HCl(aq)} + \mathrm{NaHCO}_{3}(s) \to \mathrm{NaCl(aq)} + \mathrm{H}_{2}\mathrm{O(l)} + \mathrm{CO}_{2(g)} $$ From the balanced equation, we can see that one mole of \(\mathrm{HCl}\) reacts with one mole of \(\mathrm{NaHCO}_{3}\). Since we know the number of moles of \(\mathrm{H}^{+}\) comes from \(\mathrm{HCl}\), we can assume that 0.004 moles of \(\mathrm{HCl}\) react with 0.004 moles of \(\mathrm{NaHCO}_{3}\). Now we need to convert moles of sodium hydrogen carbonate to grams, using the molar mass of \(\mathrm{NaHCO}_{3}\), which is approximately \(84\,\text{g/mol}\): $$ \mathrm{mass\ of\ NaHCO}_{3} = \mathrm{moles}\, \times \mathrm{molar\ mass} \\ \mathrm{mass\ of\ NaHCO}_{3} = 0.004\,\text{moles} \times 84\,\text{g/mol} \\ \mathrm{mass\ of\ NaHCO}_{3} = 0.336\,\text{g} $$ So, to totally neutralize the stomach acid, approximately \(0.336\,\text{g}\) of sodium hydrogen carbonate is needed.