What it the pH of 0.37M KC6H7O2(aq)? Given pKa of 4.77.

Question

Sorbic acid HC₆H₇O₂ (pka = 4.77), is widely used in the food industry as a preservative. For example it's potassium salt, (potassium sorbate) is added to cheese to inhibit the formation of mold. What it the pH of 0.37M KC₆H₇O₂(aq)?

 

 

I'm having trouble with this question since it uses pKa and I don't know what to do with it. I set up the ice table already using 

 

C6H7O2 + H2O = C6H7O2H + OH^-

 

Then I get (x^2)/(0.37-x) and thats where I get stuck

 

 

 

Answer 1

Sorbic acid is a weak acid (as indicated by the fact that it has a Ka/pKa).  Since you are looking at the salt of that weak acid (potassium sorbate), you should consider the hydrolysis, which you have done correctly except you should include the negative sign on the sorbate anion. 

 

C6H7O2^- + H2O ==> HC6H7O2 + OH^-

In this situation, the sorbate anion is acting as a base, so we can write the expression

Kb = [HC6H7O2][OH^-]/[C5H7O2^-] and we can find the Kb from the pKa since Ka x Kb = 1x10^-14 and Ka = 1.70x10^-5 (antilog pKa)

Kb = 1x10^-14/1.7x10^-5 = 5.88x10^-10.  If you prefer to calculate it another way, you can do the following:

pKa x pKb = 14.  pKb = 14 - 4.77 = 9.23.  Kb = 5.88x10^-10

5.88x10^-10 = (x)(x)/0.37 -x and if we assume that x is small relative to 0.37, we can ignore it for the time being.

x² = 2.18x10^-10

x = 1.47x10^-5 = [OH^-] (note: this is, in fact, small relative to 0.37 so our assumption is justified)

pOH = -log 1.4710^{-5 =} 4.83

pH = 14 - 4.83 = 9.17