Question

How many milliliters of 0.0695 M Ca( OH) ₂would be required to exactly neutralize 176 mL of 0.217 M HCl?

Answer 1

Normality= molarity× n-factor.

n-factor is no.of H⁺ for an acid and no.of OH^{​​​​​-} for a base.

n-factor for HCl =1

n-factor for Ca(OH)₂ =2.

Normality of Ca(OH)2 , N_{​​​​​​1} =( 0.0695 ×2) N

Normality of HCl,N_{​​​​​​2} = 0.217 N

We know

N_{​​​​​​1}​​​​​V_​1 = N_​​2​ V_​​2

  => V_{​​​​​​1} = (N₂V_{​​​​2} )÷ N_{​​​​​​1}

  => V_{​​​​​​1} = (176×0.217)÷(0.0695×2)

=274.76 mL

Answer is 274.76 mL