What volume of 0.0200 M calcium hydroxide is required to neutralize 35.00 mL of 0.0500 M nitric acid?

Short Answer

Expert verified
The volume of 0.0200 M calcium hydroxide required to neutralize 35.00 mL of 0.0500 M nitric acid is 43.75 mL.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation for the neutralization reaction between calcium hydroxide (Ca(OH)₂) and nitric acid (HNO₃) is: Ca(OH)₂ + 2HNO₃ → Ca(NO₃)₂ + 2H₂O This equation shows that 1 mole of calcium hydroxide is needed to neutralize 2 moles of nitric acid.
02

Find moles of nitric acid

To find the moles of nitric acid, we can use the formula: Moles = Molarity × Volume Given that the molarity of nitric acid is 0.0500 M and the volume is 35.00 mL, we can convert the volume to liters and find the moles: Moles of nitric acid = 0.0500 M × 0.035 L = 0.00175 moles
03

Determine moles of calcium hydroxide required

From the balanced chemical equation, we know that one mole of calcium hydroxide is needed to neutralize two moles of nitric acid, so we can determine the moles of calcium hydroxide required: Moles of Ca(OH)₂ = (0.00175 moles of HNO₃) × (1 mol Ca(OH)₂ / 2 moles HNO₃) = 0.000875 moles
04

Calculate the volume of calcium hydroxide solution

Now, we know the moles of calcium hydroxide needed and the molarity of the calcium hydroxide solution. We can use these values to find the volume required using the formula: Volume = Moles / Molarity Volume of Ca(OH)₂ = 0.000875 moles / 0.0200 M = 0.04375 L = 43.75 mL So, the volume of 0.0200 M calcium hydroxide required to neutralize 35.00 mL of 0.0500 M nitric acid is 43.75 mL.

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