How many grams of Al(OH)3 are required to neutralize 300mL of stomach acid HCl with a pH of 1.5?
pH = 1.5 = -log*(H^+)
(H^+) = 0.032 M
..............Al(OH)3 + 3HCl ==> AlCl3 + 3H2O
millimoles HCl = mL x M = 300 mL x 0.032 = 9.6
The equation tells you that 1 mol Al(OH)3 requires 3 moles NaOH; therefore, 9.6 millimoles of HCl will require 9.6/3 = 3.2 millimoles AlOH)3 or 0.0032 mols Al(OH)3. Convert to grams by
grams Al(OH)3 = mols x molar mass = ?
Post your work if you get stuck.
To determine the number of grams of Al(OH)3 required to neutralize 300 mL of stomach acid HCl with a pH of 1.5, we need to first calculate the amount of HCl in moles and then use the balanced chemical equation to determine the stoichiometric ratio between HCl and Al(OH)3.
Here's how we can do it step by step:
1. Calculate the concentration of HCl in the stomach acid:
pH is a measure of acidity, and it is calculated using the formula: pH = -log[H+]. Since the pH of the stomach acid is given as 1.5, we can calculate the concentration of H+ (hydrogen ions) using the formula: [H+] = 10^(-pH).
[H+] = 10^(-1.5) = 3.16 x 10^(-2) mol/L
2. Convert the concentration of HCl to moles:
The concentration is given in mol/L, and the volume is given in mL. We need to convert the volume to liters by dividing by 1000:
300 mL / 1000 = 0.3 L
Now we can calculate the number of moles of HCl by multiplying the concentration by volume:
Moles of HCl = concentration × volume = 3.16 x 10^(-2) mol/L × 0.3 L = 9.48 x 10^(-3) moles
3. Determine the stoichiometric ratio between HCl and Al(OH)3:
The balanced chemical equation for the neutralization reaction between HCl and Al(OH)3 is:
3HCl + Al(OH)3 → AlCl3 + 3H2O
From the equation, we can see that 3 moles of HCl react with 1 mole of Al(OH)3. Therefore, the stoichiometric ratio is 3:1.
4. Calculate the moles of Al(OH)3 required:
Since the stoichiometric ratio is 3:1, the moles of Al(OH)3 required would be one-third of the moles of HCl:
Moles of Al(OH)3 = 1/3 × Moles of HCl = 1/3 × 9.48 x 10^(-3) moles = 3.16 x 10^(-3) moles
5. Calculate the mass of Al(OH)3 required:
To calculate the mass, we need to know the molar mass of Al(OH)3, which can be calculated by adding up the atomic masses of each element: Al (26.98 g/mol) + O (16.00 g/mol) + 3H (3.01 g/mol) = 78.01 g/mol.
Mass of Al(OH)3 = moles × molar mass = 3.16 x 10^(-3) moles × 78.01 g/mol = 0.248 g
Therefore, approximately 0.248 grams of Al(OH)3 are required to neutralize 300 mL of stomach acid HCl with a pH of 1.5.
To calculate the grams of Al(OH)3 required to neutralize 300 mL of stomach acid HCl with a pH of 1.5, we need to follow several steps:
Step 1: Determine the concentration of HCl in moles per liter (M).
Since the pH of the stomach acid is given as 1.5, we can convert it to the concentration of HCl using the formula:
pH = -log[H+]
1.5 = -log[H+]
[H+] = 10^(-pH)
[H+] = 10^(-1.5)
[H+] = 0.0316 M
Step 2: Determine the balanced chemical equation between Al(OH)3 and HCl.
The balanced equation is:
Al(OH)3 + 3HCl → AlCl3 + 3H2O
Step 3: Determine the stoichiometry between Al(OH)3 and HCl.
For every 3 moles of HCl, we need 1 mole of Al(OH)3. Thus, the stoichiometric ratio is 1:3.
Step 4: Convert the volume of stomach acid to moles of HCl.
We know that 1 L of solution is equivalent to 1,000 mL. Therefore, the volume in liters is 300 mL / 1000 mL/L = 0.3 L.
moles of HCl = concentration of HCl (M) × volume (L)
moles of HCl = 0.0316 M × 0.3 L = 0.00948 moles
Step 5: Calculate the moles of Al(OH)3 required.
Using the stoichiometric ratio from step 3, we can calculate the moles of Al(OH)3.
moles of Al(OH)3 = moles of HCl × (1 mole Al(OH)3 / 3 moles HCl)
moles of Al(OH)3 = 0.00948 moles × (1 mole Al(OH)3 / 3 moles HCl)
moles of Al(OH)3 = 0.00316 moles
Step 6: Calculate the grams of Al(OH)3 required.
The molar mass of Al(OH)3 is 78 g/mol.
grams of Al(OH)3 = moles of Al(OH)3 × molar mass of Al(OH)3
grams of Al(OH)3 = 0.00316 moles × 78 g/mol
grams of Al(OH)3 = 0.24648 g
Therefore, approximately 0.24648 grams of Al(OH)3 are required to neutralize 300 mL of stomach acid with a pH of 1.5.