## Lena772 one year ago Calculate the following values for a 1.25 L solution that is 0.002915 M NaOH.

1. Lena772

[H3O+],[OH-], pOH, pH

2. Lena772

Is it NaOH+H2O--> OH- + Na+ + H2O? or NaOH + H2O-> H3O+ + NaO

3. 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

4. aaronq

it's $$\sf NaOH\rightarrow Na^++OH^-$$

5. Lena772

Its strong, but I tried 0 for H3O+ and that was wrong.

6. aaronq

it can't be 0 for $$[H_3O^+]$$, it will just be very small

7. Lena772

When I calculate X, that's what I get

8. aaronq

Use the relations: $$\sf pOH=-log[OH^-]$$ $$\sf pH=-log[H_3O^+]$$ $$\sf pH+pOH=14$$ $$\sf [OH^-][H_3O^+]=K_w=1.0*10^{-14}$$

9. Lena772

Yes but my Ka chart says the Kb of OH is 1*10^-14

10. Lena772

Based on that, H3O+ and pH should be 0, but it isn't. :/

11. aaronq

the Kb of NaOH is very very large, so that's not right

12. Lena772

Is the Kb 1 then?

13. aaronq

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

14. Lena772

So -log(0.002915) = OH-? I can do the rest from there.

15. aaronq

yes

16. Lena772

So is H3O+ 3.47e-12? Is that small enough?

17. Lena772

I got pH and pOH right but when I take 10^(-pOH) and 10^(-pH), and those are wrong so idk...

18. Lena772

It's asking for the concentrations at equilibrium but idk if that makes a difference @aaronq

19. aaronq

what did you get wrong?