Question

Calculate the pH at which bromocresol green indicator would be 50% in its yellow form and 50% in its blue form?

this question is confusing me. please help

Answer 1

pH = 4.9

Explanation:

Many indicators are weak acids which dissociate:

`sf(underbrace(HInd)rightleftharpoonsH^++underbrace(Ind^-))`

`sf(color(yellow)(Yellow)color(white)(xxxxxx)color(blue)(blue))`

`sf(K_a=([H^+][Ind^-])/([HInd])=1.3xx10^(-5))`

The dissociated form has a different colour to the molecular form.

When we have 50% of each form we can say:

`sf([H^+]=[Ind^-]`

`:.` `sf(K_a=([H^+]cancel([Ind^-]))/(cancel([HInd])))`

`sf(K_a=[H^+]=1.3xx10^(-5))`

`sf(pH=-log[H^+]=-log[1.3xx10^(-5)]=4.9)`

This is the pH at the end point of the titration. You want this to correspond to the equivalence point where the acid and base are in the molar ratio as specified by the equation.