The balanced chemical equation for the neutralization reaction between nitric acid (HNO3) and aluminum hydroxide (Al(OH)3) is:
2 Al(OH)3 + 3 HNO3 → Al(NO3)3 + 6 H2O
From the equation, we can see that 2 moles of Al(OH)3 are required to neutralize 3 moles of HNO3.
First, we need to calculate the number of moles of HNO3 in the 55.0 mL of 0.454 M solution:
moles of HNO3 = volume of solution (in L) × concentration of HNO3
moles of HNO3 = 0.0550 L × 0.454 mol/L
moles of HNO3 = 0.0250 mol
To neutralize this amount of HNO3, we need:
moles of Al(OH)3 = (2/3) × moles of HNO3
moles of Al(OH)3 = (2/3) × 0.0250 mol
moles of Al(OH)3 = 0.0167 mol
Finally, we need to calculate the volume of the aluminum hydroxide solution required to provide 0.0167 moles of Al(OH)3. The molar concentration of the aluminum hydroxide solution is not given in the problem, so we cannot directly calculate the volume of the solution. However, we can use the fact that the number of moles of a solute is equal to its concentration multiplied by its volume:
moles of Al(OH)3 = volume of Al(OH)3 solution (in L) × concentration of Al(OH)3
0.0167 mol = volume of Al(OH)3 solution (in L) × concentration of Al(OH)3
Since we do not know the concentration of the aluminum hydroxide solution, we cannot directly calculate its volume. However, we can make an assumption that the aluminum hydroxide solution is very dilute, so its concentration can be approximated as the same as the number of moles of Al(OH)3 required. This assumption is valid if the aluminum hydroxide solution is prepared by dissolving aluminum hydroxide in excess water.
Using this assumption, we can calculate the volume of the aluminum hydroxide solution:
volume of Al(OH)3 solution = moles of Al(OH)3 / concentration of Al(OH)3
volume of Al(OH)3 solution = 0.0167 mol / 0.0167 mol/L
volume of Al(OH)3 solution = 1.00 L
Therefore, we need to add 1.00 L of a very dilute aluminum hydroxide solution to completely neutralize 55.0 mL of 0.454 M HNO3