A patient is suspected of having low stomach acid, a condition known as hypochlo

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 18.00 mL sample of her gastric juices and titrate the sample with 0.000494 M KOH. The gastric juice sample required 5.27 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.

Please show all work. Thanks!

Explanation / Answer

Balanced chemical equation is:
OH- + H+ ---> H2O +


Here:
M(OH-)=4.94*10^-4 M
V(OH-)=5.27 mL
V(H+)=18.0 mL

According to balanced reaction:
number of mol of OH- =number of mol of H+
M(OH-)*V(OH-) =M(H+)*V(H+)
4.94*10^-4*5.27 = M(H+)*18.0
M(H+) = 1.0*10^-4 M

use:
pH = -log [H+]
= -log (1.0*10^-4)
= 4.0

Answer: 4.0

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