Question

A 20.0 mL sample of gastric juice was taken from a patient suspected of having hypochloridia (low stomach acid). The sample was taken several hours after a meal so there was no ingested food or drink, and it was assumed no buffers were present. The sample required 4.7 mL of 0.0019 M NaOH to titrate the sample to neutrality. What was the pH of the gastric juice?

Answer 1

To calculate the pH of the gastric juice first we calculate the moles of acid

HCl and NaOH react in a 1:1 mole ratio: NaOH + HCl ==> H2O + NaCl

mmoles NaOH = Molarity NaOH x mL NaOH = (0.0019)(4.7 ) = 0.00893mmoles NaOH

Now calculate the moles of HCl as follows:

0.00893 mmoles NaOH x (1 mmole HCl / 1 mmole NaOH) = 0.00893 mmoles HCl

Then calculate the moles of H+ as follows:

0.00893 mmoles HCl x (1 mmole H+ / 1 mmole HCl) = 0.00893 mmoles H+

Molarity H+ = mmoles H+ / mL of sample = 0.00893 / 20.0 = 0.000 4465 M H+

pH = -log [H+] = -log 0.0004465 = 3.35

Hence the pH of the gastric juice is 3.35.