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 15.00 mL sample of her gastric juices and titrate the sample with 3.99×10−4 M KOH. The gastric juice sample required 9.23 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.
To determine the pH of the gastric juice sample, we need to find the concentration of H+ ions in the solution. Since KOH is a strong base and HCl is a strong acid in the stomach, the reaction that occurs during the titration can be written as:
HCl + KOH → KCl + H2O
From the balanced equation, we can see that 1 mole of HCl (hydrochloric acid) reacts with 1 mole of KOH (potassium hydroxide) to produce 1 mole of water. Therefore, the number of moles of HCl in the gastric juice sample can be calculated using the equation:
moles of HCl = moles of KOH
The volume of KOH used in the titration is 9.23 mL, and its concentration is 3.99×10^(-4) M. We can calculate the number of moles of KOH by multiplying the volume with the concentration:
moles of KOH = (9.23 mL) × (3.99×10^-4 M) = 3.68177×10^(-6) moles of KOH
Since the reaction is 1:1 between HCl and KOH, the moles of HCl in the gastric juice sample is also equal to 3.68177×10^(-6) moles.
Now, we need to calculate the concentration of HCl in the gastric juice sample. The volume of the gastric juice sample is 15.00 mL, so the concentration of HCl can be determined as follows:
[HCl] = moles of HCl / volume of sample
= (3.68177 × 10^(-6) moles) / (15.00 × 10^(-3) L)
= 2.45451 × 10^(-4) M
Since the concentration of HCl gives the concentration of H+ ions, we can use this value to calculate the pH of the solution using the equation:
pH = -log[H+]
pH = -log(2.45451 × 10^(-4))
= -(-3.6117)
Therefore, the pH of the gastric juice sample is approximately 3.61.
To calculate the pH of the gastric juice sample, we need to determine the amount of acid neutralized by the KOH titrant.
First, let's calculate the moles of KOH used in the titration. We can use the formula:
moles of KOH = concentration of KOH x volume of KOH (in liters)
moles of KOH = (3.99×10^(-4) M) x (9.23 mL) x (1 L / 1000 mL)
moles of KOH ≈ 3.68x10^(-7) mol
Considering the balanced chemical equation for the reaction of KOH and HCl (a common component of gastric juice):
KOH + HCl -> KCl + H2O
we can determine that 1 mole of KOH reacts with 1 mole of HCl.
Therefore, the moles of HCl neutralized by KOH is also 3.68x10^(-7) mol.
Since 1 mole of HCl produces 1 mole of H+ ions in solution, the moles of H+ in the gastric juice sample is also 3.68x10^(-7) mol.
Now, let's calculate the concentration of hydrogen ions of the gastric juice sample. We can use the formula:
concentration of H+ = moles of H+ / volume of sample solution (in liters)
concentration of H+ = (3.68x10^(-7) mol) / (15.00 mL) x (1 L / 1000 mL)
concentration of H+ ≈ 2.45x10^(-5) M
Finally, we can calculate the pH of the gastric juice sample using the formula:
pH = -log(concentration of H+)
pH = -log(2.45x10^(-5))
pH ≈ 4.61
Therefore, the pH of the gastric juice sample is approximately 4.61.