How many mL of 1 M HNO3 are needed to neutralize 500 mL of 2.0 M Ca(OH)2?

Question

HNO₃****

Ca(OH)₂********

Answer 1

Neutralization of Acid and Base.

Balanced Equation: 2 HNO₃ + Ca(OH)₂ --> Ca(NO₃)₂ + 2 H₂O

Stoichiometric 2 mol 1 mol

C₁V₁ = C₂V₂

C_HNO3V_HNO3 = 2 x (C_Ca(OH)2V_ca(OH)2)

(1M)V_HNO3 = 2 x 2.0M x 500mL/1000mL

V_HNO3 = (2 x 2.0 M x 0.500 L) / 1 M



= 2.0 L or 2000 mL // answer

Answer 2

Both HNO₃ and Ca(OH)₂ are strong acid/base so they completely dissociate in solution.

First you need to find the amount of OH^- that needs to be neutralized:

2.0 M Ca(OH)₂ * 500 mL Ca(OH)₂ * [2 mol OH / 1 mol Ca(OH)₂] = 2 mol OH^-

Therefore, you need 2 mol H⁺ to neutralize the solution:

2 mol H⁺ * [1 mol HNO₃ / 1 mol H⁺] * [1 L HNO₃ / 1 mol HNO₃] = 2.0 L of the 1 M HNO₃

*** Note that in the third term in the step above is the inverse of the given HNO₃ molarity. Sometimes inverting units in stoichiometry is useful and completely fine to do so.***