Question

Identify which component of the equilibrium reaction was added↑ or removed↓ as a result of the applied stress.

CH3COOH (aq) + H2O(l) ⇆ H3O+ (aq) + CH3COO- (aq)

after added methyl orange (as indicator), color change to reddish orange.

then we split into two test tube, one as reference, other one we have to add 5 drops of 2M (NaCH3COO)

Question:

1. Identify which component of the equilibrium reaction was added↑ or removed↓ as a result of the applied stress.

2. As a result of the shift, which reaction species changed, resulting in the observed color change?

Indicate whether it increased↑ or decreased↓.

3. Give the ion(s) present in the equilibrium mixture causing the observed color:

really need help!! let me know if need more info.

Answer 1

CH3COOH(aq) + H2O(l) <------> H3O⁺(aq) + CH3COO^-(aq)

1. CH3COO^- is added

NaCH3COO is an strong electrolyte and completely dissociated into Na⁺ and CH3COO^-. So, addition of NaCH3COO means addition of CH3COO^-.

2. H3O⁺ is changed which results in colour change

H3O⁺ is decreased

CH3COO^- is conjucate base , so it reacts with H3O⁺ to give CH3COOH and water.

3. H-MeO and MeO^- (HIn and In- ) are the responsible for the observed colour

An indicator generally represented as HIn and the conjucate base of the indicator is In^-

The equillibrium between HIn and In^- is as follows

HIn(aq) + H2O(l) <------> H3O⁺(aq) + In^-(aq)

Ka = [In^-] [H3O⁺] / [ HIn ]

Henderson - Hasselbalch equation is

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

where,

pKa = pKa of the weak acid

A^- = conjucate base

HA = weak acid

From this equation we know that

at equal concentration of HIn and In^-, pH will be equal to pKa of HIn

pH > pKa , In^- will be predominate

pH < pKa , HIn will be predominate

pKa of methyl orange = 3.4

Therefore

below pH 3.4 , red colour of H-MeO will dominate

abive pH 3.4 , yellow colour of MeO^- will dominate

Working range of methyl orange is 3.1 to 4.4