How would you rearrange the Henderson–Hasselbalch equation to find out [a/ha] from pH=pKa + log[a/ha] ?

Question

Naomi Aronson · Accepted Answer

Here's how you would do that.

The Hendeson-Hasselbalch equation's most popular version enables you to determine the pH of a buffer solution containing a weak acid and its conjugate base. #color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))" "# Here #pK_a# is equal to #color(blue)(pK_a = - log(K_a))" "#, where #K_a# - the acid dissociation constant of the weak acid. That means that for a general weak acid-conjugate base buffer #"HA"_text((aq]) + "H"_3"O"_text((l]) rightleftharpoons "H"_3"O"_text((aq])^(+) + "A"_text((aq])^(-)# The solution's pH will be #"pH" = pK_a + log( (["A"^(-)])/(["HA"]))# Now, in order to determine the ratio that exists between the concentration of the conjugate base, #"A"^(-)#, and the concentration of the weak acid, #"HA"#, you will need to isolate the log term on one side of the equation #log( (["A"^(-)])/(["HA"])) = "pH" - pK_a# Now, can say that if #x = y#, then #10^x = 10^y# Consequently, the equation above will be equal to #10^(log( (["A"^(-)])/(["HA"]))) = 10^("pH" - pK_a)# However, given that #color(blue)(10^(log_10(x)) = x)# What you'll ultimately have #color(green)((["A"^(-)])/(["HA"]) = 10^("pH" - pK_a))#

Wyatt Adkins · Answer

undefined To rearrange the Henderson-Hasselbalch equation to find [A/HA] from pH = pKa + log[A/HA], you would do the following:

[A/HA] = 10^(pH - pKa)