What volume of .0500 M calcium hydroxide is required to neutralize 38.50 mL of .0400 M nitric acid?
1. Write the equation.
2. M HNO3 x L HNO3 = mols HNO3.
3. Use the equation to convert mols HNO3 to mols Ca(OH)2.
4. Now use M x L = mols to calculate volume (L) Ca(OH)2. I worked a similar example with NaOH and H2SO4 a few question above this one by the same author.
Ca(OH)2 + 2HNO3 --> 2HOH + Ca(NO3)2
(.0400 mol HNO3 / 1000mL)(38.50 mL) (1 mol Ca(OH)2 / 2 mol HNO3) (1000 mL / .0500 mol Ca(OH)2) = 1.54e1 mL
right.
Yes, you are correct. Here is the step-by-step explanation:
1. Write the balanced equation:
Ca(OH)2 + 2HNO3 --> 2H2O + Ca(NO3)2
2. Use the equation to calculate the number of moles of HNO3:
(0.0400 mol HNO3 / 1000 mL) * 38.50 mL = 0.00154 mol HNO3
3. Use stoichiometry to convert moles of HNO3 to moles of Ca(OH)2:
(0.00154 mol HNO3) * (1 mol Ca(OH)2 / 2 mol HNO3) = 0.00077 mol Ca(OH)2
4. Finally, use the equation Molarity (M) x Volume (L) = moles to calculate the volume of Ca(OH)2:
(0.0500 M Ca(OH)2) * Volume (L) = 0.00077 mol Ca(OH)2
Volume (L) = 0.00077 mol Ca(OH)2 / 0.0500 M Ca(OH)2 = 0.0154 L
Converting the volume from liters to milliliters:
0.0154 L * 1000 mL/L = 15.4 mL
So, the volume of 0.0500 M calcium hydroxide required to neutralize 38.50 mL of 0.0400 M nitric acid is 15.4 mL.