Question

A patient is suspected of having low stomach acid, a condition known as hypochloridia. To determine whether the patient as this condition, her doctors take a 15,00 mL sample of her gastric juices and titrate the sample with 1.87 X 10M KOH. The gastric juice sample required 3.42 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 pH = For the patient to be suffering from hypochloridia, the pH of the gastric juices from the stomach must be greater than pH 4. Does the patient have hypochloridia? Ono O unable to determine O yes

Answer 1

Here is the Solution:

To solve this question we will use M1 X V1 = M2 X V2 where, M1 = Molarity of HCl = ? V1= volume of HCl = 15.00 mL M₂= Molarity of KOH = 1087 x 10-4 V2² volume of KOH = 3.42 mL. Now, solve the equation for Mi. M M1 = Max V2 Na My = (1.87 X 10-4 M )* (3.42 ML) 15.00 mL Ma = 4.2636 x 10-5 M ion (H+] of The M₂ 4.2636x105 m is the hydronium HCI , Now, we will use pH formula - pH = pH = pH = -log [H +] -log (4.2636x10-5) 4, 31

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