Question

What is the volume of 5.5 M HCl solution if it is neutralized in a titration by 85.0 mL of 1.22 M Mg(OH)2?

Answer 1

Answer:

0.38dm³ or 38mL

Explanation:

Given parameters:

molarity of HCl = 5.5M

volume of Mg(OH)₂  = 85mL

molarity of Mg(OH)₂ = 1.22M

Unknown:

Volume of HCl solution = ?

Solution:

we have to adopt the mole concept to solve this problem.

  Complete reaction equation:

     2HCl   +   Mg(OH)₂   →   MgCl₂   +  2H₂O

For a standard acid titrated against a standard base,

           aA + bB → cC + dD

 $\frac{number of moles of A}{number of moles of B} = \frac{a}{b}$

A is the acid, B is the base

since number of moles = C x V

 C is the molarity

 V is the volume

 

     CₐVₐ / CₓVₓ  = a / x

 a = acid

 x = base

    85mL to dm³ = 0.085dm³

       since the unknown is Vₐ,

let us make it the subject of the expression;

             Vₐ =  (a/x)   x   ( CₓVₓ/ Cₐ)

              Va = $\frac{2}{1} x \frac{1.22 x 0.085}{5.5}$   = 0.38dm³