Calculating pH and pOH for 0.048 M Benzoic Acid Solution

[Coco12]

Homework Statement

what is the H, Oh , ph and poh for a 0.048 mol/l solution of benzoic acid? C6H5COOH

Homework Equations

Ph +poh=14
Ph=-log(h30+)
H30=10^-ph

The Attempt at a Solution

I thought maybe u need to separate benzoic acid into its ions and then use the molar ratio to find the concentration of H? but what does that break down into?

 

[Borek]

C₆H₅COOH <-> C₆H₅COO^- + H⁺

Note: it is a weak acid, so you need to use its dissociation constant to calculate pH.

 

[Coco12]

Ok we did not learn that yet but I will look it up in my chemistry book. Thanks

 

[epenguin]

Ok we did not learn that yet but I will look it up in my chemistry book. Thanks

Better still, if you haven't done it already, try and work it out without looking it up in your chemistry book. You know something about equilibrium equations and mass conservation. If you succeed you will gain confidence and understanding, if you don't succeed the explanation will mean more and be better learned when you do look it up and see maybe what key point you'd missed. If you've looked it up already, apply this principle in future problems in chemistry or any other science as far as possible. It will even make things easier for you in the long run.