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Chemistry (Acids and Bases)

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THIS HAS TWO PARTS

NaOH(s) was added to 1.0 L of HCl(aq) 0.42 M.

1) Calculate [H3O+] in the solution after the addition of 0.13 mol of NaOH(s).

2) Calculate [H3O+] in the solution after the addition of 0.81 mol of NaOH(s).

  • Chemistry (Acids and Bases) - ,

    These are best done by making an ICE chart.
    1L x 0.42M HCl = 0.42 moles HCl
    NaOH = 0.13 moles added.
    ...........HCl + NaOH ==> NaCl + H2O
    begin......0.42...0.........0......0
    add..............0.13................
    react....-0.13..-0.13....+0.13...+0.13
    final......0.29....0.......-.13...0.13

    You can see that you have 0.29 mole HCl and it's in 1L soln; therefore, HCl being a strong acid (100% ionized), the (H3O^+) = (HCl) = 0.29moles/L = 0.29M.

    The second one is done the same way except that the NaOH will be in excess. After you find the (OH^-), convert to (H3O^+) by (H3O^+)(OH^-) = Kw = 1E-14

  • Chemistry (Acids and Bases) - ,

    thank you very much

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