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Lena772

  • one year ago

Calculate the following values for a 1.25 L solution that is 0.002915 M NaOH.

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  1. Lena772
    • one year ago
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    [H3O+],[OH-], pOH, pH

  2. Lena772
    • one year ago
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    Is it NaOH+H2O--> OH- + Na+ + H2O? or NaOH + H2O-> H3O+ + NaO

  3. aaronq
    • one year ago
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    First identify whether or not NaOH is a strong electrolyte. If it is, then we can assume that it completely dissociates, that is, any ion produced is equal to the concentration of the whole compound. If it's a weak electrolyte, then you have use an equilibrium expression. Next write the dissociation equation for NaOH

  4. aaronq
    • one year ago
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    it's \(\sf NaOH\rightarrow Na^++OH^-\)

  5. Lena772
    • one year ago
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    Its strong, but I tried 0 for H3O+ and that was wrong.

  6. aaronq
    • one year ago
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    it can't be 0 for \([H_3O^+]\), it will just be very small

  7. Lena772
    • one year ago
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    When I calculate X, that's what I get

  8. aaronq
    • one year ago
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    Use the relations: \(\sf pOH=-log[OH^-]\) \(\sf pH=-log[H_3O^+]\) \(\sf pH+pOH=14\) \(\sf [OH^-][H_3O^+]=K_w=1.0*10^{-14}\)

  9. Lena772
    • one year ago
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    Yes but my Ka chart says the Kb of OH is 1*10^-14

  10. Lena772
    • one year ago
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    Based on that, H3O+ and pH should be 0, but it isn't. :/

  11. aaronq
    • one year ago
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    the Kb of NaOH is very very large, so that's not right

  12. Lena772
    • one year ago
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    Is the Kb 1 then?

  13. aaronq
    • one year ago
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    no the Kb is like in the billions, which means it's a strong (electrolyte) base, it dissociates to essentially 100% therefore \(\sf [OH^-]=[NaOH]\)

  14. Lena772
    • one year ago
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    So -log(0.002915) = OH-? I can do the rest from there.

  15. aaronq
    • one year ago
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    yes

  16. Lena772
    • one year ago
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    So is H3O+ 3.47e-12? Is that small enough?

  17. Lena772
    • one year ago
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    I got pH and pOH right but when I take 10^(-pOH) and 10^(-pH), and those are wrong so idk...

  18. Lena772
    • one year ago
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    It's asking for the concentrations at equilibrium but idk if that makes a difference @aaronq

  19. aaronq
    • one year ago
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    what did you get wrong?

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