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Simple pH problem

  1. Apr 1, 2007 #1
    I'm trying to do this homework problem, and I really am confused about how to do it. I'm familiar with dissociation of weak acids/bases, but I'm pretty confused with this problem(with a strong base and strong acid). The problem is:
    "40.0ml of 0.20M HCl is mixed with 40.5ml of 0.20M NaOH. a) Write the overall reaction and the net ionic reaction. b) Determine the [H+],[OH-],pH,pOH of the resulting solution"

    I wrote the overall and net ionic reactions, HCl + NaOH -> H2O + NaCl and H+ + OH- -> H2O, which were easy. However, when it comes to determining the pH, I'm confused. Would the H+ and OH- ions cancel out in the solution if they are present in equal ammounts(i.e. the only thing that would matter would be the .5ml excess NaOH)? I tried writing out an ICE(Initial, Change, Equilibrium) table, but I was REALLY confused about whether to use the net ionic reaction or not, and what to place in for the change. Any help would be appriciated with this problem.

    Thanks,
    Mike
     
  2. jcsd
  3. Apr 1, 2007 #2

    symbolipoint

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    As long as you have no weak acid or weak base involved, the problem description presents nothing to confuse you. Simply all of the HCl will be neutralized, and an excess of hydroxide will be present. You can safely ignore any dissociation of water for most of the work.

    Moles of OH- = 0.5 * 0.2

    [OH-] = (0.5)*(0.2)/(40 + 40.5)

    From that, you can obtain pOH, pH, and [H+]
     
  4. Apr 1, 2007 #3
    Wouldn't the moles of OH- be .0005 * .2 because the .5 is in mililiters?
     
  5. Apr 1, 2007 #4

    symbolipoint

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    mikesown,
    You are right. I made a mistake. As long as I multiplied by milliliters, I should have indicated millimoles of hydroxide. I then compensated in my message by dividing by the resulting milliliters of solution.
     
  6. Apr 3, 2007 #5

    Borek

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