What percent of #"HIn"# (#"pH"# range of #2.60#) is in the buffer solution in order to see only the acidic color of the indicator, if the acid-base ratio to see the acidic color is equal to the base-acid ratio to see the basic color?

I got #95.2%# #"HIn"#.

If #(["HIn"])/(["In"^(-)])# to see the acidic color is equal to #(["In"^(-)])/(["Hin"])# to see the basic color, whatever ratio that is, they are separate scenarios, but their quantities are the same so that the distance of both pH boundaries are equidistant from the #"pK"_a#.

And so, using both versions of the Henderson-Hasselbalch equation,

#"pH"_b = "pK"_a + log((["In"^(-)])/(["Hin"]))# #" "" "" "" "bb((1))#

#14 - "pH"_a = "pK"_b + log((["HIn"])/(["In"^(-)]))# #" "" "bb((2))#

where #"pH"_a# is the minimum pH in the range, and #"pH"_b# is the maximum #"pH"# in the range. In each case, the ratio of acid to base is equal to the other, but is not #1#.

Upon solving #(1)# and #(2)# for #"pK"_a#, we obtain (using #"pK"_a + "pK"_b = 14#):

#"pK"_a = "pH"_b - log((["In"^(-)])/(["Hin"]))# #" "" "" "bb((3))#

#"pK"_a = "pH"_a + log((["HIn"])/(["In"^(-)]))# #" "" "" "bb((4))#

Setting #(3)# equal to #(4)# gives:

#"pH"_b - log((["In"^(-)])/(["Hin"])) = "pH"_a + log((["HIn"])/(["In"^(-)]))#

#Delta"pH"_(ab) = "pH"_b - "pH"_a = 2.60#

#= log((["In"^(-)])/(["Hin"])) + log((["HIn"])/(["In"^(-)]))#

Since the acid/base ratios are numerically equal in different scenarios, call the ratio #c_(ab)#.

#2.60 = 2log(c_(ab))#,

#=> c_(ab) = 10^(1.30) = 19.95#

So, if #19.95 = (["HIn"])/(["In"^(-)])# (as it must be to see only #"HIn"#), then

#19.95["In"^(-)] = ["HIn"]# #" "" "" "" "" "bb((5))#

and that means #["HIn"]# is #bb19.95# times more present than #["In"^(-)]#.

We know that

#%"HIn" + %"In"^(-) = 100%#,#" "" "bb((6))#

so since #%"In"^(-) cdot 19.95 = %"HIn"# from #(5)#, plug that into #(6)# to get:

#%"HIn" + (%"HIn")/(19.95) = 100%#.

Solving this, we get:

#%"HIn"(1 + 1/19.95) = 100%#

#=> color(blue)(%"HIn") = (100/(1 + 1/19.95))% = color(blue)ul(95.2%)#

to see only the acidic color.

What percent of #"In"^(-)# is required to see only the basic color?

EXAMPLE

Methyl red would be an example with a #"pH"# range of #4.40 - 6.20#, or #1.80# #"pH"# units. It turns red below #"pH"# #4.40# ("acidic", dominated by #"HIn"#), and yellow above #"pH"# #6.20# ("basic", dominated by #"In"^(-)#).

That is, the acidic #"HIn"# color is seen below #"pH"# #4.40#, and the basic #"In"^(-)# color is seen above #"pH"# #6.20#.

Using the methyl red example for a reality check, we have that its #"pK"_a# is #4.95#.

That means that from #(1)# and #(2)#:

#4.95 = 6.20 - log((["In"^(-)])/(["Hin"]))#

#4.95 = 4.40 + log((["HIn"])/(["In"^(-)]))#

In this case, it's an example where the ratios of acid and base are not equal in either case.

See if you can show that #color(green)(%"In"^(-) = 94.7%)# to see the basic (yellow) color at #color(green)("pH" = 6.20)#, and that #color(green)(%"HIn" = 78.0%)# to see the acidic (red) color at #color(green)("pH" = 4.95)#.