Chem II

Calculate the pH of the solution made by mixing 35.0 mL of 0.500 M ca(OH)2 with 45.0 mL of 0.710 MHCl

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asked by Ken
  1. Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O

    mols Ca(OH)2 = M x L = 0.500M x 0.035L = 0.0175 mols.
    mols HCl = M x L = 0.710M x 0.045L = 0.03195

    Since the reaction is 1 mol Ca(OH)2 to 2 mols HCl, that means if all the Ca(OH)2 reacts, it will require 0.0175 x (2 mol HCl/1 mol Ca(OH)2) = 0.0175 x 2/1 = 0.035 mols HCl and we don't have that much; therefore, the HCl is the limiting reagent, all of it will react, and some of the Ca(OH)2 will remain behind unreacted. How much will that be.
    We do the reverse calculation for HCl.
    0.03195 mols HCl x (1 mol Ca(OH)2/2 mols HCl) = 0.01598 mols Ca(OH)2 required. So how much is left.
    0.01750-0.01598 = 0.00153 mols Ca(OH)2.
    What is the volume will we have.
    35.0 mL + 45.0 mL = 80.0 mL = 0.080 L so
    [Ca(OH)2] = 0.00153/0.080 = 0.01906 M. Since there are two OH^- per Ca(OH)2 molecule, (OH^-) is twice that or 0.01906 x 2 = ??
    pOH form that.
    pH from that.
    Post your work if you get stuck.
    Check my arithmetic. These numbers all look alike on my keyboard.

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