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PH and POH question

  1. Jun 20, 2011 #1
    1. The problem statement, all variables and given/known data

    NaOH is an example of a strong base. if 0.45g of NaOH is mixed with 500ml of water what is the


    2. Relevant equations

    3. The attempt at a solution

    so, i calculating moles thru molar mass and divided by 0.5 to get mol/l. put that as the concentration of the hydroxyl. (0.0225)

    What i dont understand is how to get the hydronium ion concentration. I know that its going to involve doing the -log of 0.025 to get the pH. Which is 1.4. But why do i do 14-1.4 then do the antilog of that to get the concentration of the hydronium ion?

    I know how its solved, My real question is... why do i do 14- (ph of hydroxl) to find the concentration of the hydronium?
  2. jcsd
  3. Jun 20, 2011 #2
    Because water's autohydronolysis has the equilibrium constant of 10^-14
    [OH-]*[H3O]=10^-14 <--> pOH- + pH = 14

    WHen you pour a base into it, a negligible part of that base will be converted to water until the equilibrium condition is met
  4. Jun 20, 2011 #3


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    Staff: Mentor

    More datils: water ion product.
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