## willowdavis97 2 years ago What is the pH of a solution that has a concentration of hydronium ions that is 1.8 × 10-4 M OH-? Show, or explain, the work used to solve this problem.

1. willowdavis97

makes no sense I have no idea how to do this

2. aaronq

pOH = -log[OH-] pH + pOH =14

3. aaronq

you meant hydroxide ions, right?

4. willowdavis97

I don't so that was the exact question.

5. willowdavis97

Is there a formula? I could do it if I had one.

6. aaronq

in your question, you said hydronium but then you wrote OH-, which one is it?

7. willowdavis97

I think the (-) is supposed to be up higher if that makes a difference. 1.8 × 10^-4 M OH^-

8. willowdavis97

I guess that makes no sense at all

9. aaronq

lol no i understand that part.. but hydronium = H3O+ and hydroxide OH- so it's hydroxide? lol i'm just confused b/c you said two different things and one or the other completely changes the answer

10. willowdavis97

Well the question it asked me was that so i dont know if its right or not :(

11. aaronq

there must be a type somewhere if it's 1.8 × 10^-4 M OH^- .. use the equations i gave you first if it's 1.8 × 10^-4 M H3O+ then just use pH= -log[H3O+]

12. aaronq

typo**

13. willowdavis97

what do I put into the equation. Whats a -log Ahhh sorry I sound so stupid I suck so bad at chemistry :(

14. aaronq

you input the number.. 1.8 × 10^-4 or 0.00018 then press log on your calculator .. which turns into a logarithm base 10 then press the negative sign

15. aaronq

lol it's cool, everyone's gotta learn someday, no one is born with this knowledge

16. willowdavis97

okay I did that I got - 3.74472749..... Does that sound right?

17. aaronq

you didn't press the negative sign, it's supposed to be 3.74 if it was a negative number it would be very very acidic

18. willowdavis97

Ohhhh, okay. Does that mean the ph is 3.74..?

19. aaronq

if the concentration they gave you was in H3O+ ions, if it was OH- you have to subtract from 14

20. willowdavis97

okay so 14-3.74? so 10.26

21. aaronq

yep

22. willowdavis97

Thank you!

23. aaronq

no problem !