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 20.00 mL20.00 mL sample of her gastric juices and titrate the sample with 4.82×10−4 M KOH.4.82×10−4 M KOH. The gastric juice sample required 1.31 mL1.31 mL of the KOHKOH 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.

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Asked Sep 7, 2019
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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 20.00 mL20.00 mL sample of her gastric juices and titrate the sample with 4.82×10−4 M KOH.4.82×10−4 M KOH. The gastric juice sample required 1.31 mL1.31 mL of the KOHKOH 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.

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Expert Answer

Step 1

The main constituent of Gastric juice is hydrochloric acid (stomach acid).

Given,

Volume of HCl (V1) = 20.00 mL = 0.020 L

Molarity of KOH (M2) = 4.82 × 10-4 M

Volume of KOH (V2) = 1.31 mL = 0.00131 L

Step 2

The reaction taking place can be written as:

HCl + KOH ------->  KCl + H2O

From the above reaction, it is evident that 1 mole of HCl reacts with 1 mole of KOH. At neutral point, number of moles of HCl will be equal to number of moles of KOH.

The molarity of HCl (M1) can be calculated as:

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At neutral point: M1 Vi M2 x V2 4.82 x 104 M x 0.00131 L M1 0.020 L M1 3.16 x 10-5 M

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Step 3

HCl is a strong acid and it dissociates completely to give H+ and Cl...

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HСІ H*Cl [HCI] H [Cl] 3.16 x 10-5 M pH log [H pH log (3.16 x 105) pH 4.5

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