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 17.00 mL sample of her gastric juices and titrate the sample with 0.000463 M KOH. The gastric juice sample required 1.60 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.

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?

A. Yes

B. no

C. Unable to determine

Answer 1

Gastric juice contains HCl, so write the reaction of HCl with KOH

HCl + KOH -----> KCl + H₂O

Calculate the moles of KOH required 1 to neutralize gastric juices

Moles of KOH = Molarity of KOH x Volume of KOH in L

Moles of KOH = 0.000463 M x 1.60 mL /1000 = 7.41 x 10^-7 moles

From equation, 1 mol KOH neutralizes 1 mol HCl

So, 7.41 x 10^-7 moles KOH neutralizes 7.41 x 10^-7 mol HCl

Now calculate molarity of HCl

Molarity of HCl = mol of HCl / Volume of HCl in L

Molarity of HCl =7.41 x 10^-7 /17.00 mL /1000 = 4.36 x 10^-5 M

Now calculate the PH of 4.36 x 10^-5 M HCl

HCl is strong acid, so it dissociates completely.

[H⁺] = 4.36 x 10^-5

PH = - log [H⁺]

PH = -log (4.36 x 10^-5) = 4.36

For the patient to be suffering from hypochloridia, the pH of the gastric juices from the stomach must be greater than pH 4. In this case PH of patient’s gastric juices is 4.36 which is greater than pH 4.So the patient has hypochloridia.

Option A is correct.