An acid-base indicator which is a weak acid has a pK a value = 5.45 . At what concentration ratio of sodium acetate to acetic acid would the indicator show a colour half-way between those of its aci

Question

An acid-base indicator which is a weak acid has a pK a ​ value = 5.45 . At what concentration ratio of sodium acetate to acetic acid would the indicator show a colour half-way between those of its acid and conjugate base forms ? [ pK a ​ of acetic acid = 4.75]

TestHub · Accepted Answer

Explanation:

The question asks for the concentration ratio of sodium acetate (the conjugate base) to acetic acid (the weak acid) when an indicator shows a colour halfway between its acid and base forms. This occurs when the pH of the solution equals the pK_aof the indicator.

Given Data:

pK_aof the indicator = 5.45

pK_aof acetic acid = 4.75

Concept:

When an indicator is halfway between its acid and base forms, the pH of the solution is equal to its pK_{a .}

The Henderson-Hasselbalch equation relates pH, pK_a, and the ratio of conjugate base to acid:

$p H = p K_{a} + lo g (\frac{[ co nj ug a t e ba se ]}{[ a c i d ]})$

Calculation:

1. Since the indicator is halfway between its forms, the pH of the solution is equal to the indicator's pK_a, which is 5.45.

2. We want to find the ratio of $[so d i u ma ce t a t e] / [a ce t i c a c i d]$ .

3. We can use the Henderson-Hasselbalch equation and rearrange it:

$p H = p K_{a} + lo g (\frac{[ so d i u ma ce t a t e ]}{[ a ce t i c a c i d ]})$

$5.45 = 4.75 + lo g (\frac{[ so d i u ma ce t a t e ]}{[ a ce t i c a c i d ]})$

$lo g (\frac{[ so d i u ma ce t a t e ]}{[ a ce t i c a c i d ]}) = 5.45 - 4.75 = 0.7$

$\frac{[ so d i u ma ce t a t e ]}{[ a ce t i c a c i d ]} = 1 0^{0.7} \approx 5.01$

4. Thus, the ratio is approximately 5:1.

Answer: option 3, (5:1)

Chemistry - Ionic Equilibrium Question with Solution | TestHub

Options:

Doubts & Discussion

JEE Main 2028 (Class 11)