Molaritybase*volumebase*Heq=Molarityacid*volumeacid*Heq
Molaritybase*14.9ml*3=1.75*22.75ml*3
molaritybase= you do it...
Molaritybase*14.9ml*3=1.75*22.75ml*3
molaritybase= you do it...
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.
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.