Lena772
  • Lena772
Calculate the following values for a 1.25 L solution that is 0.002915 M NaOH.
Chemistry
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katieb
  • katieb
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Lena772
  • Lena772
[H3O+],[OH-], pOH, pH
Lena772
  • Lena772
Is it NaOH+H2O--> OH- + Na+ + H2O? or NaOH + H2O-> H3O+ + NaO
aaronq
  • aaronq
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

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aaronq
  • aaronq
it's \(\sf NaOH\rightarrow Na^++OH^-\)
Lena772
  • Lena772
Its strong, but I tried 0 for H3O+ and that was wrong.
aaronq
  • aaronq
it can't be 0 for \([H_3O^+]\), it will just be very small
Lena772
  • Lena772
When I calculate X, that's what I get
aaronq
  • aaronq
Use the relations: \(\sf pOH=-log[OH^-]\) \(\sf pH=-log[H_3O^+]\) \(\sf pH+pOH=14\) \(\sf [OH^-][H_3O^+]=K_w=1.0*10^{-14}\)
Lena772
  • Lena772
Yes but my Ka chart says the Kb of OH is 1*10^-14
Lena772
  • Lena772
Based on that, H3O+ and pH should be 0, but it isn't. :/
aaronq
  • aaronq
the Kb of NaOH is very very large, so that's not right
Lena772
  • Lena772
Is the Kb 1 then?
aaronq
  • aaronq
no the Kb is like in the billions, which means it's a strong (electrolyte) base, it dissociates to essentially 100% therefore \(\sf [OH^-]=[NaOH]\)
Lena772
  • Lena772
So -log(0.002915) = OH-? I can do the rest from there.
aaronq
  • aaronq
yes
Lena772
  • Lena772
So is H3O+ 3.47e-12? Is that small enough?
Lena772
  • Lena772
I got pH and pOH right but when I take 10^(-pOH) and 10^(-pH), and those are wrong so idk...
Lena772
  • Lena772
It's asking for the concentrations at equilibrium but idk if that makes a difference @aaronq
aaronq
  • aaronq
what did you get wrong?

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