Final answer:
To neutralize 34.7 mL of 0.775 M NaOH and 58.5 mL of 0.0100 M Al(OH)3, 269 mL and 17.55 mL of 0.100 M HNO3 are needed, respectively, calculated by using mole ratios from balanced chemical equations and formulas.
Step-by-step explanation:
The question involves the calculation of volumes of 0.100 M HNO₃ needed to neutralize two different solutions: one containing NaOH and the other containing Al(OH)₃. To solve this, we need to use the concept of titration, which is a technique used to determine the concentration of a solution by reacting it with a solution of known concentration.
For the first solution, we have 34.7 mL of 0.775 M NaOH. The balanced equation for the reaction with HNO₃ is:
HNO₃ + NaOH -> NaNO₃ + H₂O
This indicates that the reaction ratio between HNO₃ and NaOH is 1:1.
First, calculate the moles of NaOH:
Moles of NaOH = 34.7 mL x 0.775 M x (1 L / 1000 mL) = 0.0269 mol
Since the ratio is 1:1, moles of HNO₃ needed = moles of NaOH = 0.0269 mol
Calculate the volume of HNO₃ required:
Volume of HNO₃ = Moles / Concentration = 0.0269 mol / 0.100 M = 0.269 L = 269 mL
For the second solution with 58.5 mL of 0.0100 M Al(OH)₃, the balanced equation is:
3HNO₃ + Al(OH)₃ -> Al(NO₃)₃ + 3H₂O
The mole ratio is 3:1 (HNO₃:Al(OH)₃). Calculate moles of Al(OH)₃:
Moles of Al(OH)₃ = 58.5 mL x 0.0100 M x (1 L / 1000 mL) = 0.000585 mol
Since the ratio is 3:1, moles of HNO₃ needed = 3 x moles of Al(OH)₃ = 3 x 0.000585 mol = 0.001755 mol
Calculate the volume of HNO₃ required:
Volume of HNO₃ = Moles / Concentration
Volume of HNO₃ = 0.001755 mol / 0.100 M = 0.01755 L = 17.55 mL