what mass of aluminum hydroxide needs to be consumed to neutralize 1.25L of stomach acid (HCl). Stomach acid has a pH of 2.2
pH = -log(H^+)
-2.2 = log(H^+) and
(H^+) = 0.00631 M
molsl H^+ = M x L = 0.00631 x 1.25 L = aboaut 0.008 but you need to that more accurately.
Al(OH)3 + 3HCl = AlCl3 + 3H2O
mols HCl = about 0.008
Convert mols HCl to mols Al(OH)3 using the coefficients in the balanced equation. That will give you 0.008 x 1/3 = mols Al(OH)3.
Convert that to grams. g = mols x molar mass.
To determine the mass of aluminum hydroxide needed to neutralize the 1.25L of stomach acid, we need to follow a series of steps:
Step 1: Write the balanced equation for the neutralization reaction between aluminum hydroxide (Al(OH)3) and hydrochloric acid (HCl).
2 Al(OH)3 + 6 HCl -> 2 AlCl3 + 6 H2O
Step 2: Determine the molar ratio between aluminum hydroxide and hydrochloric acid.
From the balanced equation, we can see that 2 moles of aluminum hydroxide react with 6 moles of hydrochloric acid. Therefore, the molar ratio is 2:6, which can be simplified to 1:3.
Step 3: Calculate the number of moles of hydrochloric acid.
To do this, we need to convert the volume of stomach acid from liters to moles. We can use the formula:
moles = volume (L) × concentration (mol/L)
Given that the pH of the stomach acid is 2.2, we can assume it has a concentration of 10^(-2.2) mol/L.
moles of HCl = 1.25 L × (10^(-2.2) mol/L) = 1.25 × 10^(-2.2) mol
Step 4: Calculate the number of moles of aluminum hydroxide required.
Using the molar ratio from step 2, we can determine the moles of aluminum hydroxide needed:
moles of Al(OH)3 = moles of HCl × (1 mole Al(OH)3 / 3 moles HCl)
moles of Al(OH)3 = (1.25 × 10^(-2.2) mol) × (1 mole Al(OH)3 / 3 moles HCl)
Step 5: Calculate the mass of aluminum hydroxide.
Finally, we can use the molar mass of aluminum hydroxide (Al(OH)3) to calculate the mass required. The molar mass of Al(OH)3 is 78 g/mol.
mass of Al(OH)3 = moles of Al(OH)3 × molar mass of Al(OH)3
mass of Al(OH)3 = [ (1.25 × 10^(-2.2) mol) × (1 mole Al(OH)3 / 3 moles HCl) ] × (78 g/mol)
By performing this calculation, you will find the mass of aluminum hydroxide required to neutralize 1.25L of stomach acid (HCl).