Question

What volume of 0.100 m HNO₃ is required to neutralize 58.5 ml of 0.0100 m al(OH)3?

Answer 1

Approximately 1.94 mL of 0.100 M HNO₃ is required to neutralize 58.5 mL of 0.0100 M Al(OH)₃.

Step-by-step explanation:

To determine the volume of 0.100 M HNO₃ required to neutralize 58.5 mL of 0.0100 M Al(OH)₃, we can use the concept of mole ratio. The balanced chemical equation for the reaction between HNO₃ and Al(OH)₃ is 3HNO₃ + Al(OH)₃ → Al(NO₃)₃ + 3H₂O. From the equation, we can see that for every 3 moles of HNO₃, 1 mole of Al(OH)₃ is neutralized. Therefore, we can set up a proportion to solve for the volume:

(0.100 M HNO₃) / (x mL HNO₃) = (0.0100 M Al(OH)₃) / (58.5 mL Al(OH)₃)

Solving for x, we find that approximately 1.94 mL of 0.100 M HNO₃ is required.

Answer 2

To neutralize 58.5 mL of 0.0100 M Al(OH)₃, approximately 17.55 mL of 0.100 M HNO₃ is required.

I hope this is right :)