calcium hydroxide is titrated with nitric acid.What volume of 0.0200 M calcium hydroxide is required to neutralize 35.00 ml of 0.0500 M nitric acid?

2HNO3 + Ca(OH)2 ==> Ca(NO3)2 + 2H2O

35.00 mL x 0.0500 M = 1.750 mmols HNO3. That will require 1/2 x 1.750 = mmoles Ca(OH)2.
M = mmoles/mL
You have M Ca(OH)2 and mmoles Ca(OH)2, solve for mL Ca(OH)2.

0.875/0.0200 = 43.75mL

To find the volume of 0.0200 M calcium hydroxide (Ca(OH)2) required to neutralize 35.00 mL of 0.0500 M nitric acid (HNO3), we can use the concept of stoichiometry. In the balanced chemical equation for the reaction between calcium hydroxide and nitric acid, the stoichiometric ratio shows that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

Here's how we can solve it step by step:

Step 1: Write the balanced chemical equation for the reaction between calcium hydroxide and nitric acid:
2 HNO3 + Ca(OH)2 -> Ca(NO3)2 + 2 H2O

Step 2: Find the number of moles of nitric acid:
Molarity (M) = Moles (n) / Volume (V)
Rearranging the equation, we can find the number of moles:
Moles (n) = Molarity (M) x Volume (V)
Moles of nitric acid = 0.0500 M x (35.00 mL / 1000 mL/1 L)
= 0.00175 moles

Step 3: Since the stoichiometric ratio between HNO3 and Ca(OH)2 is 2:1, we know that 2 moles of HNO3 react with 1 mole of Ca(OH)2.

Step 4: Calculate the number of moles of calcium hydroxide required:
Moles of Ca(OH)2 = 0.00175 moles of HNO3 / 2
= 0.000875 moles

Step 5: Find the volume of 0.0200 M calcium hydroxide solution using the mole-to-volume relationship:
Moles (n) = Molarity (M) x Volume (V)
Rearranging the equation, we can find the volume:
Volume (V) = Moles (n) / Molarity (M)
Volume of calcium hydroxide (V) = 0.000875 moles / 0.0200 M
= 0.04375 L x (1000 mL/1 L)
= 43.75 mL

Therefore, the volume of 0.0200 M calcium hydroxide required to neutralize 35.00 mL of 0.0500 M nitric acid is 43.75 mL.

438mL