Question
What volume of .300M HCL is required (a) to titrate 50.0 ml of 0.600 M LiOH? (B) to titrate 50.0ml of .600M Ca(OH)2
(balanced equation and solution for both please!)
(balanced equation and solution for both please!)
Answers
DrBob222
I'll do the Ca(OH)2 but the process is the same for LiOH.
Ca(OH)2 + 2HCl ==> 2H2O + CaCl2
step 1. Write and balance the equation.
step 2. Convert what you have, in this case Ca(OH)2), to mols. mols = M x L = 0.600 x 0.050 = 0.03 mols Ca(OH)2.
step 3. Using the coefficients in the balanced equation, convert mols of what you have [Ca(OH)2] to mols of what you want [HCl].
0.03 mols Ca(OH)2 x [2 mol HCl/1 mol Ca(OH)2] = 0.03 x 2/1 = 0.06 mols HCl. [Note how the factor convert mols Ca(OH)2 to mols HCl by canceling unit of mols Ca(OH)2 and but leaving unit of HCl].
step 4. Now you have mols HCl.
M HCl = mols HCl/L HCl. You have 0.3M HCl and mols HCl = 0.06, solve for L.
L = mols/M = 0.06/0.3 = 0.2L = 200 mL.
Ca(OH)2 + 2HCl ==> 2H2O + CaCl2
step 1. Write and balance the equation.
step 2. Convert what you have, in this case Ca(OH)2), to mols. mols = M x L = 0.600 x 0.050 = 0.03 mols Ca(OH)2.
step 3. Using the coefficients in the balanced equation, convert mols of what you have [Ca(OH)2] to mols of what you want [HCl].
0.03 mols Ca(OH)2 x [2 mol HCl/1 mol Ca(OH)2] = 0.03 x 2/1 = 0.06 mols HCl. [Note how the factor convert mols Ca(OH)2 to mols HCl by canceling unit of mols Ca(OH)2 and but leaving unit of HCl].
step 4. Now you have mols HCl.
M HCl = mols HCl/L HCl. You have 0.3M HCl and mols HCl = 0.06, solve for L.
L = mols/M = 0.06/0.3 = 0.2L = 200 mL.