Question

What is the molarity of the Ca(OH)2 solution if 32.00 mL of Ca(OH)2 requires 16.08 mL of a 2.303 M solution for complete titration?

Answer 1

Answer:

For this problem we just need to remember the equation and that the volume is always in liters: MaVa=MbVb

  Ma= 1.338 mol/L Va= 18.75 mL= 0.01875 L Mb= x Vb=24.73 mL= 0.02473 L

 

  So now we can plug into the equation and solve:

   1.338 mol/L * 0.01875 L= x mol/L * 0.02473 L

This is a two step process: stoichiometry and using the answer from the first part to plug and chug it into the Molariy equation.

  First step: setup the stoichiometry problem:

  3.1171 g Na2CO3 * (1 mol/106 g) * (2 mol HCl/ 1 mol Na2CO3)= mol HCl

 

  Second step: Molarity equation

  Molarity= moles HCl/ L   M= mol HCl/0.04027 L

For the third problem, you will just use the same equation as the first: MaVa= MbVb

  Ma=0.57 M   Va= x L   Mb= 0.875 M  Vb= 23.83 mL= 0.02383 L

  0.57 M * x L= 0.875 M * 0.02383 L

With this equation, we want to find the moles of NaOH by using the molarity equation first then because MaVa= MbVb, we know that the number of moles has to be equal.

   0.75 M= x mol/ 0.0227 L    mol NaOH = 0.75 M *0.0227 L         mol NaOH= 0.017025

  So next, we can set it to the mass of the acid using this equation: Molar Mass= mass/ moles

  Molar Mass= 3.6 g/ 0.017025 mol  

 And with that you will find the molar mass of HX, and even determine what X is.

Explanation:

hope I helped