Chemistry

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work out the [OH-], [H+] and pH of the solution formed when 0.500L of 0.250 M hydrochloric acid solution is mixed with 0.500 L of a 0.250 M calcium hydroxide solution.

  • Chemistry -

    I think it is easier to set up an ICE chart for these. Note that it is a limiting reagent problem.
    mol HCl = M x L = 0.5 x 0.25 = 0.125
    mol Ca(OH)2 = 0.5 x 0.25 = 0.125

    .......2HCl + Ca(OH)2 ==> CaCl2 + 2H2O
    I.....0.125....0.125......0........0
    C....-0.125...-0.0625...0.0625..0.0625
    E......0.......0.0625...0.0625..0.0625

    Therefore, the final solution is one in which you have 0.0625 mol Ca(OH)2 in 1L of solution.
    You can do OH, pOH, pH from there?

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