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  • one year ago

Calculate either [H3O ] or [OH–] for the solutions below at 25 °C. [OH-]=1.83x10^-7 M [H3O+]=? M

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  1. Photon336
    • one year ago
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    They are asking for the concentration of hydronium ions H3O+. Hydronium ion's just mean's H+. Here's he formula that you use. \[[H3O ^{+}][OH ^{-}] = 1x10^{-14} = Kw @25C\] I believe that this is called the ion product constant; kw for water. like any equilibrium constant, kw,ksp,kp etc.. it's going to depend on temperature. so at 25 degrees celsius this number will always be the same for water. what you need to do is this tells you that the product of the concentrations = 1x10^-14 at that temperature. so to find [H3O] you just need to play around with the equation: \[\frac{ 1x10^{-14} }{ [OH] } = [H3O ^{+}]\] THE [ ] means concentration moles/liter another interesting thing is that you can get the pH too p just means take the -log of the concentration [ ] so it would be -log[H3O+] that's if they asked you for that. FYI even better you can also get the pOH too which is just the - log of the concentration of your OH- ions. \[pH + pOH = 14 \] \[14 - pH = pOH\] So in summary the interesting thing with these problems is if you know one number then you can easily get the other because the formulas equal a constant. FYI if your'e interested (NOT RELATED TO YOUR PROBLEM) as to where we got pH + pOH = 14 let's look at this formula' [H3O][OH] = 1x10^-14 forgive me i'm not the best at math here, but you take the -log of both sides. (Hopefully a math expert can spot any possible errors in this). \[-\log( [OH][H3O] ) = -Log(1x10^{-14})\] \[-\log(xy) = -\log(x) + -\log(y) = 14 \] substitute, \[-\log = p\] \[pOH + pH = 14\]

  2. Photon336
    • one year ago
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