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 6.3 mL of 0.0011 M NaOH to titrate the sample to neutrality.

What was the pH of the gastric juice?

Answer 1

mols NaOH = M x L = ?
mols HCl(in the gastric juice) = mols NaOH
M HCl = mols HCl/L HCl. L = 0.020.
Then convert M to pH.
pH = -log(H^+).

Answer 2

-1.70

Answer 3

12