Question

You are measuring 2,6 dichlorophenol in an aqueous sample. The pKa of this material is 6.79 degrees C in water, so it is a weak acid. The log Kow for it is 4.00. You decide to extract it from the aqueous sample using an organic solvent but need to adjust the pH to make the process effective and efficient. Solve for or indicate the pH required for 99.9% of the 2,6 dichlorophenol to be in an extractable form.  

Answer 1

Distribution ratio (D) = E/(100-E), where E= extracted or extractable form of 2,6 dichlorophenol=99.9%

here,D=99.9/(100-99.9)=99.9/0.1=999 therefore D=999

Again,   D=k_ow[H₃O⁺]/(k_a+[H₃O⁺])  where,k_ow=partition co-efficient ,[H₃O⁺]=concentration of hydronium ion or proton and k_a=dissociation constant for weak acid here 2,6 dichlorophenol

so, k_ow[H₃O⁺]/(k_a+[H₃O⁺])=999

here,logk_ow=4.00 so,k_ow=10⁴​ and pKa=6.79 so,k_a=10^-6.79

therefore,10⁴[H₃O⁺]/ (10^-6.79 +[H₃O⁺])=999

or,[H₃O⁺]/ (10^-6.79 +[H₃O⁺])=999/10⁴ =0.0999

or,[H₃O⁺]=0.0999(110^-6.79+[H₃O⁺])

or,(1-0.0999)[H₃O⁺]=0.0999×10^-6.79

or, [H₃O⁺]=1.8×10^-8

or, pH=7.74

So, the required pH is 7.74

+[H₃O⁺])