A patient is suspected of having low stomach acid, a condition known as hypochloridia. To determine whether the patient has this condition, her doctors take a 15.00 mL sample of her gastric juices and titrate the sample with 3.99×10−4 M KOH. The gastric juice sample required 9.23 mL of the KOH titrant to neutralize it.

Question

Calculate the pH of the gastric juice sample. Assume the sample contained no ingested food or drink which might otherwise interfere with the titration.

Answer 1

To calculate the pH of the gastric juice sample, we need to determine the amount of acid neutralized by the KOH titrant.

First, let's calculate the moles of KOH used in the titration. We can use the formula:

moles of KOH = concentration of KOH x volume of KOH (in liters)

moles of KOH = (3.99×10^(-4) M) x (9.23 mL) x (1 L / 1000 mL)
moles of KOH ≈ 3.68x10^(-7) mol

Considering the balanced chemical equation for the reaction of KOH and HCl (a common component of gastric juice):

KOH + HCl -> KCl + H2O

we can determine that 1 mole of KOH reacts with 1 mole of HCl.

Therefore, the moles of HCl neutralized by KOH is also 3.68x10^(-7) mol.

Since 1 mole of HCl produces 1 mole of H+ ions in solution, the moles of H+ in the gastric juice sample is also 3.68x10^(-7) mol.

Now, let's calculate the concentration of hydrogen ions of the gastric juice sample. We can use the formula:

concentration of H+ = moles of H+ / volume of sample solution (in liters)

concentration of H+ = (3.68x10^(-7) mol) / (15.00 mL) x (1 L / 1000 mL)
concentration of H+ ≈ 2.45x10^(-5) M

Finally, we can calculate the pH of the gastric juice sample using the formula:

pH = -log(concentration of H+)

pH = -log(2.45x10^(-5))
pH ≈ 4.61

Therefore, the pH of the gastric juice sample is approximately 4.61.