How many milliliters of 0.500 M HBr would be required to react with 40.0 mL of 0.300 M Ca(OH)2?

48dm³

Explanation:

Given reaction:

                  Ca(OH)₂ + 2HBr →  CaBr₂ + 2H₂O

Parameters:

Concentration of HBr = 0.5M

 Volume of Ca(OH)₂ = 40mL

 Concentration of  Ca(OH)₂ = 0.3M

Solution:

To solve this problem, we are going to use the mole concept. We solve from the known specie to the unknown.

 We first find the number of moles of the known specie which is the Ca(OH)₂ ;

                number of moles  = concentration x volume

                number of moles  = 0.3 x 40 x 10⁻³ = 0.012moles

From the reaction equation;

           1 mole of Ca(OH)₂  requires 2 moles HBr

          0.012 moles of Ca(OH)₂ will require 0.012 x 2  = 0.024moles of HBr

Now,

 To find the volume of HBr;

          Volume  =

          Volume  =   = 0.048dm³

In mL;

       Volume 0.048 x 1000 = 48dm³

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.