How would you use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.120 M in HClO and 0.185 M in KClO?

1 Answer
Nov 22, 2015

Answer:

#"pH" = 7.65#

Explanation:

The Henderson - Hasselbalch equation allows you to calculate the pH of buffer solution that contains a weak acid and its conjugate base by using the concentrations of these two species and the #pK_a# of the weak acid.

#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))#

In your case, the weak acid is hypochlorous acid, #"HClO"#. Its conjugate base, the hypochlorite anion, #"ClO"^(-)#, is delivered to the solution by one of its salts, potassium hypochlorite, #"KClO"#.

The acid dissociation constant, #K_a#, for hypochlorous acid is equal to #3.5 * 10^(-8)#

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Now, before doing any calculations, try to predict what you expect the solution's pH to be compared with the acid's #pK_a#.

Notice that when you have equal concentrations of weak acid and conjugate base, the log term will be equal to zero, since

#log(1) = 0#

This tells you that if you have more conjugate base than weak acid, the log term will be greater than #1#, which will cause the pH to be higher than the #pK_a#.

With this in mind, plug in your values into the H-H equation to get

#"pH" = -log(3.5 * 10^(-8)) + log( (0.185color(red)(cancel(color(black)("M"))))/(0.120color(red)(cancel(color(black)("M")))))#

#"pH" = 7.46 + 0.188 = color(green)(7.65)#