A student adds 15.0 mL 2.00 M Acetic Acid and 15.0 mL 2.00 M NaOH. Determine the pH of the resulting solution.

Ka HC2H3O2 = 1.80E-5

Question

A student adds 15.0 mL 2.00 M Acetic Acid and 15.0 mL 2.00 M NaOH. Determine the pH of the resulting solution.

Ka HC2H3O2 = 1.80E-5

anonymous · Answer

This was my thought process. But it's saying that I have the answer incorrect by more than 10%

aaronq · Answer

you have to use the concentrations as equilibrium, after the reaction (f any) has occured. The NaOH will neutralize the acetic acid, and you'll be left with 0.03 moles of acetate in 30 mL. so can just use the Kb and an eq. expression to find the pH

anonymous · Answer

@aaronq  Okay, so would the equilibrium expression be Kc= [acetate] [OH-]/ [acetic acid]??

aaronq · Answer

nope,    $Acetate+H_2O\rightleftharpoons acetic~ acid+OH^-$

 $K_B=\dfrac{[acetic~ acid][OH^-]}{[acetate]}$

aaronq · Answer

oh, and $\sf K_B=\dfrac{K_w}{K_A}$