My brain refuses to wrap around titration. Problem: Given 150.0 mL of a 1.05 M HCl solution, how many milliliters of a 2.50 M Al(OH)3 solution are required to neutralize the HCl?
I think the balanced formula is 3HCl + Al(OH)3 - AlCl +3H20 but that's as far as I get.
To solve this titration problem, you need to use stoichiometry and the concept of molarity. Here's a step-by-step explanation to help you solve it:
Step 1: Write the balanced chemical equation:
The balanced equation you provided is incorrect. The correct balanced equation for the reaction between HCl and Al(OH)3 is:
3HCl + Al(OH)3 → AlCl3 + 3H2O
Step 2: Calculate the moles of HCl:
Moles = Molarity × Volume (in liters)
Given that the volume of the HCl solution is 150.0 mL (which is equal to 0.1500 L) and the concentration is 1.05 M, you can calculate the moles of HCl:
Moles of HCl = 1.05 mol/L × 0.1500 L = 0.1575 moles
Step 3: Use stoichiometry to find the moles of Al(OH)3:
From the balanced equation, you can see that the ratio of HCl to Al(OH)3 is 3:1. So, the moles of Al(OH)3 will be one-third of the moles of HCl:
Moles of Al(OH)3 = 0.1575 moles × 1/3 = 0.0525 moles
Step 4: Calculate the volume of the Al(OH)3 solution:
Now that you have the moles of Al(OH)3, you can use the molarity of the Al(OH)3 solution to find the volume:
Volume (in liters) = Moles / Molarity
Volume (in liters) = 0.0525 moles / 2.50 mol/L
Volume (in liters) = 0.021 L
Step 5: Convert the volume to milliliters:
To convert the volume from liters to milliliters, multiply by 1000:
Volume (in mL) = 0.021 L × 1000 mL/L = 21 mL
Therefore, 21.0 mL of the 2.50 M Al(OH)3 solution are required to neutralize the HCl solution.