Answer: approximately 18.75 mL of 3.0 M HNO₂ solution can be completely neutralized by 75 mL of 1.5 M Mg(OH)₂ solution.
Step-by-step explanation:
To determine the volume of 3.0 M HNO₂ solution that can be neutralized by 75 mL of 1.5 M Mg(OH)₂ solution, we can set up an equation based on the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the neutralization reaction between HNO₂ and Mg(OH)₂ is:
2 HNO₂ + Mg(OH)₂ → Mg(NO₂)₂ + 2 H₂O
From the equation, we can see that it takes two moles of HNO₂ to react with one mole of Mg(OH)₂.
First, let's calculate the number of moles of Mg(OH)₂ present in 75 mL of 1.5 M solution:
Number of moles of Mg(OH)₂ = Volume (in L) × Concentration (in mol/L)
= 75 mL × (1 L / 1000 mL) × 1.5 mol/L
= 0.1125 mol
Since the stoichiometry ratio is 2:1 (HNO₂:Mg(OH)₂), we can conclude that 0.1125 moles of Mg(OH)₂ can neutralize 0.1125/2 = 0.05625 moles of HNO₂.
Now, let's determine the volume of 3.0 M HNO₂ required to contain 0.05625 moles:
Volume (in L) = Number of moles / Concentration (in mol/L)
= 0.05625 mol / 3.0 mol/L
≈ 0.01875 L
Finally, we convert the volume from liters to milliliters:
Volume (in mL) = Volume (in L) × 1000
≈ 0.01875 L × 1000
≈ 18.75 mL