If it takes 22.75 mL of 1.75 M H3PO4 to neutralize 14.90 mL of Al(OH)3, what is the molarity of the Al(OH)3?
Molaritybase*volumebase*Heq=Molarityacid*volumeacid*Heq
Molaritybase*14.9ml*3=1.75*22.75ml*3
molaritybase= you do it...
To find the molarity of Al(OH)3, we need to use the stoichiometry of the balanced chemical equation between H3PO4 and Al(OH)3 to calculate the number of moles of Al(OH)3.
The balanced chemical equation is:
3 H3PO4 + Al(OH)3 -> AlPO4 + 3 H2O
From the balanced equation, we can see that 3 moles of H3PO4 react with 1 mole of Al(OH)3.
Step 1: Calculate the number of moles of H3PO4
Given:
Volume of H3PO4 = 22.75 mL
Molarity of H3PO4 = 1.75 M
Use the formula: Moles = Molarity * Volume (in liters)
Moles of H3PO4 = 1.75 M * (22.75 mL / 1000 mL/L) = 0.03978 moles
Step 2: Calculate the number of moles of Al(OH)3
Using the stoichiometry from the balanced equation:
1 mole of Al(OH)3 corresponds to 3 moles of H3PO4
So, Moles of Al(OH)3 = (0.03978 moles of H3PO4) / 3 = 0.01326 moles
Step 3: Calculate the molarity of Al(OH)3
Given:
Volume of Al(OH)3 = 14.90 mL
Use the formula: Molarity = Moles / Volume (in liters)
Molarity of Al(OH)3 = 0.01326 moles / (14.90 mL / 1000 mL/L) = 0.8893 M
Therefore, the molarity of Al(OH)3 is 0.8893 M.
To find the molarity of Al(OH)3, we need to use the concept of stoichiometry and the balanced chemical equation of the reaction between H3PO4 and Al(OH)3.
The balanced chemical equation for the reaction is:
3H3PO4 + Al(OH)3 → Al(H2PO4)3 + 3H2O
From the equation, we can see that it takes 3 moles of H3PO4 to react with 1 mole of Al(OH)3. Therefore, the number of moles of H3PO4 is calculated as follows:
Moles of H3PO4 = (volume of H3PO4 in liters) x (molarity of H3PO4)
= (22.75 mL / 1000 mL/L) x (1.75 mol/L)
= 0.0398125 mol
Now, using the stoichiometry of the balanced equation, we can determine the moles of Al(OH)3:
Moles of Al(OH)3 = (moles of H3PO4) / 3
= 0.0398125 mol / 3
= 0.013270833 mol
Finally, we calculate the molarity of Al(OH)3:
Molarity of Al(OH)3 = (moles of Al(OH)3) / (volume of Al(OH)3 in liters)
= (0.013270833 mol) / (14.90 mL / 1000 mL/L)
= 0.88989886 M (rounded to 5 decimal places)
Therefore, the molarity of Al(OH)3 is approximately 0.8899 M.