[Boxxxed]

 A buffer solution is prepared by dissolving 1.50 g each of benzoic acid, C6H5CO2H (MW = 122 g/mol), and sodium benzoate, NaC6H5CO2 (MW=144 g/mol) in 150 mL of solution.  Ka = 6.3x10-5

a. What is the pH of this buffer solution?  

Moles acid = 0.0123
Moles base = 0.0104

Base reacts with acid to form 0.0104 conjugate base. 0.0019 mole acid is left.

Base = 0.069 M
Acid = 0.0123 M

Acid+H2O-->Base+H3O

Common Ion

Ka=(0.069+x)(x) / 0.0123-x=6.5x10^-5

ph = 4.95

Is this correct?

[rabolisk]

No. It's a bit confusing when you say acid or base. It's better to just say benzoic acid or benzoate, or write their formulas. This would have helped you in this question. Try again.

[Boxxxed]

I did a similar text book problem and got the right answer. I wasn't supposed to react the acid and base. Now it is just confusing as to why you initially react them stoichiometrically in some questions and not in others.


For example, when a buffer is already established and we add HCl, it will disassociate completely into H3O ions that will react stoichiometrically with the buffer's base, adding the same stoichiometric amount to the buffer's conjugate acid.

What if a weak acid is added? Would'n't we need the ka of the weak acid to determine how much of it will disassociate into H3O+ ions?

Ie we add 0.1 moles weak acid

Weak Acid+H2O---->Base+H3O

Ka = x² / 0.1-x  Then we use x to add/subtract from the buffer's conjugate pair. Correct train of thought?

[Borek]

What if a weak acid is added? Would'n't we need the ka of the weak acid to determine how much of it will disassociate into H3O+ ions?

In general, if a weak acid is added, you are in troubles. Its dissociation means moving the equilibrium of the buffer, there is no simple and painless way of calculating new pH of the solution. In specific cases you can think about simplifying assumptions, but general case is a PITA.

If you add the same acid that is already present, you can assume it doesn't dissociate, it just adds to the concentration of acid already present. That's not entirely true, but in most cases error you make with this assumption can be safe neglected.