"Stomach acid" is hydrochloric acid. A sample of gastric juice having a volume of 6.68 mL required 13.90 mL of 0.0161 M KOH solution for neutralization in a titration. What was the molar concentration of HCl in this fluid?
C(HCl) =
If we assume a density of 1.00 g mL-1 for the fluid, what was the percentage by weight of HCl?
w(HCl) =
Isn't this just like the others except one is HCl and the other acetic?
To find the molar concentration of HCl in the gastric juice sample, we can use the concept of stoichiometry.
First, we need to determine the number of moles of KOH that reacted in the neutralization. This can be calculated using the volume and molarity of KOH solution used.
n(KOH) = volume(KOH) x concentration(KOH)
n(KOH) = 13.90 mL x 0.0161 M
n(KOH) = 0.2241 mmol (since the volumes are given in mL, the result is in mmol)
The balanced chemical equation for the neutralization reaction is:
HCl + KOH → KCl + H2O
From the balanced equation, we can see that 1 mole of HCl reacts with 1 mole of KOH. Therefore, the number of moles of HCl in the gastric juice sample is also 0.2241 mmol.
Now, let's calculate the molar concentration of HCl in the gastric juice sample.
Volume of gastric juice sample = 6.68 mL = 6.68 cm^3
Assuming a density of 1.00 g mL^-1, the mass of the gastric juice sample can be calculated as:
Mass(gastric juice) = volume(gastric juice) x density
Mass(gastric juice) = 6.68 g
Since the molar concentration is defined as moles of solute per liter of solution, we need to convert the volume of the gastric juice sample from mL to L:
Volume(gastric juice sample) = 6.68 mL = 0.00668 L
Now, we can calculate the molar concentration of HCl:
C(HCl) = moles(HCl) / volume(gastric juice sample)
C(HCl) = 0.2241 mmol / 0.00668 L
C(HCl) ≈ 33.59 M
Therefore, the molar concentration of HCl in the gastric juice sample is approximately 33.59 M.
To calculate the percentage by weight of HCl, we need to know the molar mass of HCl. The molar mass of HCl is calculated as:
Molar mass(HCl) = atomic mass(H) + atomic mass(Cl)
Molar mass(HCl) = 1.00794 g/mol + 35.453 g/mol
Molar mass(HCl) ≈ 36.461 g/mol
Now, we can calculate the mass of HCl in the sample:
Mass(HCl) = moles(HCl) x molar mass(HCl)
Mass(HCl) = 0.2241 mmol x 36.461 g/mol
Mass(HCl) ≈ 8.17 mg
Finally, we can calculate the percentage by weight of HCl:
w(HCl) = (Mass(HCl) / Mass(gastric juice)) x 100
w(HCl) = (8.17 mg / 6.68 g) x 100
w(HCl) ≈ 0.122%
Therefore, the percentage by weight of HCl in the gastric juice sample, assuming a density of 1.00 g mL^-1, is approximately 0.122%.