how many mL of 0.10 M sodium acetate would be needed to make the pH of the solution in part b (pH=2.85) equal to the pka of acetic acid?

Question

From the first question^ what would the pH of the solution be after adding 1.0 ml of 0.10 M NaOH

Answer 1

Perhaps Henderson Hasselbach will work:

 

pH = pKa + log[A-]/[HA]

 

HA is acetic acid, A- is acetate ion, pKa acetic acid is 4.75  

 

(pKa means the negative log of the K (equilibrium constant))

 

So...2.85 = 4.75 + log of [acetate]/[acetic acid]

 

That's as far as I can get w/o knowing the concentration of the acetic acid.  Mathematically, the term must equal negative 1.90.  So, if the concentration of the acetic acid is known, for example by knowing the original pH, then the problem can be solved.