Question

Given that the $K_a\left(methylorange\right)=4.0\times {10}^{-4}$, a solutiona at $pH=2$ containing the indicator would be :

• A. Orange
• B. Yellow
• C. Colourless
• D. Red

Answer 1

Answer: (D)

Red

Given that

$K_a=4.0\times {10}^{-4}$

$p{K}_a=3.4,pH=2$ (strong acidic)

We know that,

Methyl orange has a $pH$ range $3$ to $5$ at this $pH$ range it changes its colour.

Colour at lower $pH$- red

Colour at higher $pH$ - yellow

Thus,

Colour of methyl orange at $pH-2$ would be red.

option D