How do pH and pKa relate?

1 Answer

The #"p"H# of a solution is directly related to the #"p"K_"a"# of a solution via the Henderson-Hasselbach equation,

#"p"H = "p"K_"a" + log(([A^-])/([HA]))#

Let's do an example:

What is the #"p"H# of a #"1-L"# solution of #0.12"M"# of #NH_4Cl# to which #"1 L"# of #0.03"M"# of #NaOH# was added (#"p"K_"a"# of #NH_4^(+)# is #9.25^([1])#)?##

Consider the equilibrium,

#NH_4^(+) + OH^(-) rightleftharpoons H_2O + NH_3#

It is safe to assume that the hydroxide ion will consume one equivalent ammonium's protons, leaving #0.09"mol"# #NH_4^+# ions and #0.03"mol"# #NH_3#.

Since the total volume cancels out in the ratio of concentrations, we can translate these concentrations into mols and proceed.

To be sure, the hydroxide is treated as a strong base, and the ammonium as a weak acid.

Hence,

#"p"H = "p"K_"a" + log(([NH_3]_"eq")/([NH_4^+]_"eq"))#

#= 9.25 + log("0.03 mols"/("0.12 mols" - "0.03 mols")) approx 8.77#

[1]: Nelson, D. L., Cox, M. M., & Lehninger, A. L. (2017). Lehninger Principles of Biochemistry. New York, NY: W.H. Freeman and Company.