Chemistry

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!)

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asked by Gage
  1. 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.

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    posted by DrBob222

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