[FouRRaW]

Hi, my question is....

 Find [H3O] [OH-] and pH in 1.0 L of the following solutions of strong acids and bases:
a) 0.50 M HI
b) 0.22 M RbOH

I know the formula to find pH and the formula to find  concentration.  pH= -log Base 10  [H30 ]
sorry i don't know how to subscript with this program. 

So do i have to write out the reaction with reactants and products, and then use stoichiometry to find concentrations or what?? Cause i don't know how to write these reactions.
Maybe i just don't get the question.  Any hints?  Thanks

[syko sykes]

Since you're using strong acids/bases you can assume 100% dissociation.

For a: [H₃O]=[HI]=.5M
For b: [OH^-]=[RbOH]=.22M

that should give you enough information to get you started

to do subscripts use the button that says "sub" above the smilies and the text box

[mike]

I think as they are strong acids and bases you can assume complete dissociation so the concentration of the acid 0.50M should be the same as the concentration of the hydronium ion, right?

HI + H₂O = H₃O⁺ + I^-

[HI] = [H₃O⁺]

pH = -log₁₀[H₃O⁺]

edit: haha syko sykes is too fast for me

[Borek]

http://www.chembuddy.com/?left=pH-calculation&right=pH-strong-acid-base

http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-strong-acid-q1

[Donaldson Tan]

You might get a more accurate value if you know the Ka and Kb

[villageidiot]

No need to write an equation.  Once you've found the concentration of either H₃O⁺ or OH^-, simply solve for the missing concentration:

[H₃O⁺] * [OH^-] = K_w = 1.00 x 10^-14

(the brackets [ ] mean concentration in molarity)

[FouRRaW]

H₂O

[FouRRaW]

So thanks a lot guys, that helps, i totally missed the "strong Acids part"

How come the sub script think don't work for me? (i type in the number 2, then highlight it and then type sub, then i click preview and it just says the sub boxes

I actually got another question if that's alright.

Find   [H₃O⁺]  and [OH^-] of the following solutions of strong acids and bases: 
a) a solution prepared by dissolving 12.0 g of LiOH in water to make 250mL of solution.

attempt: so I'm just not sure if this is right....

[H₂O<sup>] + LiOH  = [OH-] + Li [H₂O]

So did i write out the reaction correctly? I figured LiOH was the strong base so i gave it a hydrogen ion.
I'm not sure where the negative ion symbol goes, are Li and H2O supposed to be separate molecules in the product?? Li [H₂O]   
I hope this makes sense, and thanks!</sup>

[syko sykes]

i think the reaction should be this:
LiOH --> Li⁺ + OH^-

can you go from there?

and for the superscripts, put [sup*] before what you want to be a superscript and [/sup*] after it... remove the *'s ofcourse
same goes for subscripts except replace "sup" with "sub"

[Borek]

[H₃O⁺] + [OH^-] = K_w = 1.00 x 10^-14

[H₃O⁺] * [OH^-] = K_w = 1.00 x 10^-14

[Borek]

How come the sub script think don't work for me? (i type in the number 2, then highlight it and then type sub, then i click preview and it just says the sub boxes

Highlit then click on ^sup or _sub icons above edit field and smileys. You may add [] tags manually as well, just remember that every tag must be closed.

[H₂O] + LiOH  = [OH^-] + Li [H₂O]

It is wrong, but syko have already pushed you in the right direction.

[villageidiot]

[H₃O⁺] * [OH^-] = K_w = 1.00 x 10^-14

Right you are Borek . . . my mistake!   
I'm going to fix my post as well.

[Borek]

Right you are Borek . . . my mistake!   

Let's call it a typo

[FouRRaW]

Whoa, slow down.  I got confused with the messages, something about someone made a mistake in the equation? 
What was the equation again?
 My guess is.... H₂O + LiOH = OH^-(H₂O)   i'm not sure where the negative ion sign goes for the products also.

[Borek]

My guess is...

...wrong.


Check reaction posted by syko.