Question

Bromophenol blue is an indicator with a `K_a=5.84xx10^(-5)`. What is the `%` of indicator in it's basic form at a `pH=4.84`? Given, `log(5.84)=0.7664` & antilog of `0.6064` is `4.04`. Thank you:)

Answer 1

`80.2%`

Explanation:

The idea here is that bromophenol blue acts as a weak acid in aqueous solution.

`"HBb"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "Bb"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)`

As you can see, this equilibrium is influenced by the `"pH"` of the solution, i.e. by the concentration of hydronium cations present in the solution.

In your case, you know that you have

`"pH" = 4.84`

This implies that you have

`["H"_3"O"^(+)] = 10^(-4.84)`

Now, by definition, the acid dissociation constant is equal to

`K_a = (["Bb"^(-)] * ["H"_3"O"^(+)])/(["HBb"])`

Rearrange to find the ratio that exists between the equilibrium concentration of the conjugate base, which is the indicator is its basic form, and the equilibrium concentration of the weak acid.

`(["Bb"^(-)])/(["HBb"]) = K_a/(["H"_3"O"^(+)])`

Plug in your values to find

`(["Bb"^(-)])/(["HBb"]) = (5.84 * 10^(-5))/10^(-4.84)`

`(["Bb"^(-)])/(["HBb"]) = 5.84 * 10^((-5 + 4.84)) = 5.84 * 10^(-0.16)" "" "color(darkorange)("(*)")`

Now, in order to find the percent dissociation of the weak acid, you need to take into account the fact the initial concentration of the weak acid is equal to

`["HBb"]_0 = ["Bb"^(-)] + ["HBb"]`

This is the case because the acid dissociates in a `1:1` mole ratio to produce the conjugate base, which implies that in order to have an equilibrium concentration of `["Bb"^(-)]` and an equilibrium concentration of `["HBb"]`, the initial concentration of the weak acid must decrease by `["Bb"^(-)]`.

This means that the percent dissociation of the weak acid will be equal to

`"% dissociation" = (["Bb"^(-)])/(["Bb"^(-)] + ["HBb"]) * 100%`

Use equation `color(darkorange)("(*)")` to say that

`["Bb"^(-)] = 5.84 * 10^(-0.16) * ["HBb"]`

The percent dissociation of the weak acid will thus be equal to

`"% dissociation" = (5.84 * 10^(-0.16) * color(red)(cancel(color(black)(["HBb"]))))/(5.84 * 10^(-0.16)color(red)(cancel(color(black)(["HBb"]))) + color(red)(cancel(color(black)(["HBb"])))) * 100%`

`"% dissociation" = (5.84 * 10^(-0.16))/(5.84 * 10^(-0.16) + 1) * 100%`

Now, you know that

`log(5.84) = 0.7664" "` and `" " 10^0.6064 = 4.04`

This means that you can write

`10^log(5.84) = 10^0.7664`

`5.84 = 10^0.7664`

Plug this into the equation to get

`"% dissociation" = (10^0.7664 * 10^(-0.16))/(10^0.7664 * 10^(-0.16) + 1) * 100%`

`"% dissociation" = 10^0.6064/(10^0.6064 + 1) * 100%`

Finally, you will end up with

`"% dissociation" = 4.04/(4.04 + 1) * 100% = color(darkgreen)(ul(color(black)(80.2%)))`

I'll leave the answer rounded to three sig figs.

Answer 2

`sf(80.2%)`

Explanation:

Indicators like bromophenol blue are weak acids where the undissociated (acidic) form is a different colour from the dissociated (basic) form:

`sf("HIn"rightleftharpoonsH^++In^-)`

`sf(color(yellow)(yellow)" "color(blue)(blue))`

`sf(K_a=([H^+][In^-])/([HIn])=5.84xx10^(-5))`

`:.` `sf(([In^-])/([HIn])=K_a/[[H^+])`

We know that `sf(pH=4.84)` `:.` `sf(-log[H^+]=4.84)` from which `sf([H^+]=1.445xx10^(-5)color(white)(x)"mol/l")`

`:.` `sf(([In^-])/([HIn])=(5.84xxcancel(10^(-5)))/(1.445xxcancel(10^(-5)))=4.048)`

To get the percentage which are dissociated we can find `sf(alpha)` the degree of dissociation, or the fraction which are dissociated from an ICE table:

`sf(" "HInrightleftharpoonsH^++In^-)`

`sf(I" "1" " " "0" " " "0)`

`sf(C" "-alpha" " +alpha" "+alpha)`

`sf(E" "(1-alpha)" "alpha" "alpha)`

We know that `sf(alpha/((1-alpha))=4.048)`

`:.` `sf(alpha=4.048(1-alpha))`

`sf(alpha=4.048-4.048alpha)`

`sf(5.048alpha=4.048)`

`sf(alpha=4.048/5.048=0.802)`

This means that the basic, dissociated form is 80.2 %.