Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

When I calculate X, that's what I get

Answer 8

Use the relations:

\(\sf pOH=-log[OH^-]\)

\(\sf pH=-log[H_3O^+]\)

\(\sf pH+pOH=14\)

\(\sf [OH^-][H_3O^+]=K_w=1.0*10^{-14}\)

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

Is the Kb 1 then?

Answer 13

no the Kb is like in the billions, which means it's a strong (electrolyte) base, it dissociates to essentially 100%

therefore \(\sf [OH^-]=[NaOH]\)

Answer 14

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

Answer 15

yes

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

what did you get wrong?