Calculate the volume of 0.128 M HCl solution required to neutralize 2.32 g of Al(OH)3(s)
{2.32 / [molar mass Al(OH)3(s)]} * 3 = moles (OH⎺)
each mole of OH⎺ needs one mole of HCl to neutralize it
The problem with this problem is the solubility of Al(OH)₃. With a Ksp = 4 x 10ˉ¹⁵, the solubility of Al(OH)₃ will be ~1.1 x 10ˉ⁴M in Al(OH)₃ dissolved (the rest will ppt) and the [OHˉ] = 3(1.1 x 10ˉ⁴)M = 3.3 x 10ˉ⁴ mol/L in OHˉ b/c Al(OH)₃ => Al⁺³ + 3OHˉ. So, the 2.32-g of Al(OH)₃ when added into 1 L solution will deliver only 3.3 x 10ˉ⁴ mole in OHˉ ions. If this is filtered and titrated, then...
(M x V) HCl = (M x V) OHˉ
(0.128M)(V-Liters) = (0.00033 mol/L OHˉ)(1 Liter)
V-Liters HCl needed = (0.00033 mol)/(0.128-mol/L) = 0.00234-L = 2.34-ml of 0.128M HCl needed.
NOTE => However, if both the ppt and solution are being titrated with HCl, then as the OH^- is converted to H2O on addition of HCl, more Al(OH)3 will dissolve into solution. Such will then require ~ 697-ml of 0.128M HCl to completely convert all the OH^- to water.
To determine the volume of 0.128 M HCl solution required to neutralize 2.32 g of Al(OH)3(s), we need to follow these steps:
1. Write the balanced chemical equation for the neutralization reaction between HCl and Al(OH)3:
Al(OH)3 + 3HCl → AlCl3 + 3H2O
2. Calculate the molar mass of Al(OH)3:
Mass of Al(OH)3 = 2.32 g
Molar mass of Al = 26.98 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of Al(OH)3 = (26.98 g/mol) + 3[(1.01 g/mol) + (16.00 g/mol)] = 78.00 g/mol
3. Calculate the number of moles of Al(OH)3:
Moles of Al(OH)3 = (2.32 g) / (78.00 g/mol)
4. According to the balanced equation, the stoichiometric ratio of Al(OH)3 to HCl is 1:3. This means that in order to neutralize Al(OH)3, three moles of HCl are required.
Moles of HCl = 3 * Moles of Al(OH)3
5. Calculate the volume of HCl solution using the molarity (0.128 M) and the number of moles of HCl:
Volume of HCl solution = (Moles of HCl) / (Molarity of HCl)
Now, let's perform the calculations:
Step 2: Molar mass of Al(OH)3
Molar mass of Al(OH)3 = (26.98 g/mol) + 3[(1.01 g/mol) + (16.00 g/mol)] = 78.00 g/mol
Step 3: Moles of Al(OH)3
Moles of Al(OH)3 = (2.32 g) / (78.00 g/mol) ≈ 0.0297 mol
Step 4: Moles of HCl
Moles of HCl = 3 * Moles of Al(OH)3 = 3 * 0.0297 mol = 0.0891 mol
Step 5: Volume of HCl solution
Volume of HCl solution = (0.0891 mol) / (0.128 mol/L) ≈ 0.696 L (rounded to three decimal places)
Therefore, approximately 0.696 liters of 0.128 M HCl solution are required to neutralize 2.32 g of Al(OH)3(s).
To calculate the volume of the HCl solution required to neutralize the Al(OH)3(s), we need to determine the number of moles of Al(OH)3 and use the balanced chemical equation to find the mole ratio between HCl and Al(OH)3. We can then use this mole ratio to calculate the volume of the HCl solution.
First, we need to determine the number of moles of Al(OH)3. To do this, we divide the mass of Al(OH)3 (2.32 g) by its molar mass. The molar mass of Al(OH)3 is the sum of the atomic masses of aluminum (Al), oxygen (O), and hydrogen (H):
Molar mass of Al(OH)3 = (1 × Al atomic mass) + (3 × O atomic mass) + (3 × H atomic mass)
Using the atomic masses from the periodic table, we calculate:
Molar mass of Al(OH)3 = (1 × 26.98 g/mol) + (3 × 16.00 g/mol) + (3 × 1.01 g/mol)
= 78.00 g/mol
Now, we can calculate the number of moles of Al(OH)3:
Number of moles = Mass / Molar mass
= 2.32 g / 78.00 g/mol
≈ 0.0297 mol
Next, we need to use the balanced chemical equation between HCl and Al(OH)3 to find the mole ratio between them. The balanced equation is:
Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(l)
From the equation, we can see that the mole ratio between Al(OH)3 and HCl is 1:3.
Finally, we can calculate the volume of the HCl solution required. Since the concentration of the HCl solution is given in molarity (0.128 M), we can use the definition of molarity to relate moles and volume:
Molarity (M) = Moles / Volume (in liters)
Rearranging the equation, we find:
Volume (in liters) = Moles / Molarity
Substituting the values, we get:
Volume (in liters) = 0.0297 mol / 0.128 mol/L
≈ 0.232 L
Therefore, the volume of the 0.128 M HCl solution required to neutralize 2.32 g of Al(OH)3(s) is approximately 0.232 L.