Question

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 16.00 mL sample of her gastric juices and titrate the sample with 3.76 x 10-4 M KOH. The
gastric juice sample required 5.87 mL of the KOH titrant to neutralize it.
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

Answer:

pH of the gastric juice

$3.86$

Explanation:

As we know,

In an acid base titration, the number of milimoles of acid is equal to the number of milimoles of base.

Milimoles $=$ Molarity $*$ Volume

Molarity of gastric juices $*$ Volume of gastric juices  $=$  Molarity of KOH $*$ Volume of KOH

Substituting the given values in above equation, we get -

$16 * X = 5.87 * 3.76 * 10^{-4}$

X represents the molarity of gastric juices

$X = \frac{5.87 * 3.76 * 10^{-4}}{16} \\X = 1.379 * 10^{-4}$

pH of gastric juices is equal to log $1.379 * 10^{-4}$

$= - log 1.379 * 10^{-4}\\= - [-3.86]\\= 3.86$