# Indicator Question (1 Viewer)

#### KoshX87

##### Member
I think any of these equations are right as long as you only have one and that one contains both. If you don't have water as a reactant or as a product then you need to put aqueous and that suggests that water is present. The equation I used for this was:

Ind-(aq) + H+(aq) <-> HInd(aq) (I'm not too sure about the state of HInd, but i think aq is right)

I wrote that added acid increases [H+] and so by LCP, the equilibrium will shift to the right to form more HInd which would make the colour of the solution more red.
If an acid was added then the OH- ions would neutralise H+ and so the [H+] ions would decrease and by LCP, the equilibrium would shift left to produce more H+ and also Ind-. This would cause the solution to get more green and less red.

I concluded the question and said: Because of this, the solution can be used as an acid(red) and base(green) indicator.

#### Menomaths

##### Exaı̸̸̸̸̸̸̸̸lted Member
I think any of these equations are right as long as you only have one and that one contains both. If you don't have water as a reactant or as a product then you need to put aqueous and that suggests that water is present. The equation I used for this was:

Ind-(aq) + H+(aq) <-> HInd(aq) (I'm not too sure about the state of HInd, but i think aq is right)

I wrote that added acid increases [H+] and so by LCP, the equilibrium will shift to the right to form more HInd which would make the colour of the solution more red.
If an acid was added then the OH- ions would neutralise H+ and so the [H+] ions would decrease and by LCP, the equilibrium would shift left to produce more H+ and also Ind-. This would cause the solution to get more green and less red.

I concluded the question and said: Because of this, the solution can be used as an acid(red) and base(green) indicator.
Perfect 4/3

(Only because I mentioned everything you mentioned + same equation)

#### Infntie

##### Member
Indicators are typically weak acids which in turn account for the equilibrium that was established (i.e ionises in water to still form HInd and Ind-). Then I wrote the rest which has been clearly discussed above.