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!)
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.
To determine the volume of 0.300 M HCl required to titrate both 50.0 mL of 0.600 M LiOH and 50.0 mL of 0.600 M Ca(OH)2, we need to use the balanced equations and perform stoichiometric calculations.
a) Titrating 50.0 mL of 0.600 M LiOH with 0.300 M HCl:
Balanced equation: 2LiOH + HCl -> Li2O + 2H2O
From the balanced equation, we can see that 2 moles of LiOH react with 1 mole of HCl. Therefore, the stoichiometric ratio between LiOH and HCl is 2:1.
To determine the volume of HCl required, we can use the following formula:
Volume (LiOH) x Concentration (LiOH) = Volume (HCl) x Concentration (HCl)
Let's plug in the given values:
50.0 mL (LiOH) x 0.600 M (LiOH) = Volume (HCl) x 0.300 M (HCl)
Volume (HCl) = (50.0 mL x 0.600 M) / 0.300 M
Volume (HCl) = 100.0 mL
Therefore, 100.0 mL of 0.300 M HCl is required to titrate 50.0 mL of 0.600 M LiOH.
b) Titrating 50.0 mL of 0.600 M Ca(OH)2 with 0.300 M HCl:
Balanced equation: Ca(OH)2 + 2HCl -> CaCl2 + 2H2O
From the balanced equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of HCl. Therefore, the stoichiometric ratio between Ca(OH)2 and HCl is 1:2.
Using the same formula as above:
Volume (Ca(OH)2) x Concentration (Ca(OH)2) = Volume (HCl) x Concentration (HCl)
Plug in the given values:
50.0 mL (Ca(OH)2) x 0.600 M (Ca(OH)2) = Volume (HCl) x 0.300 M (HCl)
Volume (HCl) = (50.0 mL x 0.600 M) / (0.300 M x 2)
Volume (HCl) = 50.0 mL
Therefore, 50.0 mL of 0.300 M HCl is required to titrate 50.0 mL of 0.600 M Ca(OH)2.