Question

radish extract indicator turns to a bright red color in low ph solutions and mustard-yellow in solutions with a high ph value.
if a teacher places this radish extract indicator in a naoh water solution, which of the following accurately explains what will happen?

the indicator will turn bright red because there are more h⁺ than oh⁻ ions, and the solution is acidic.

the indicator will turn mustard-yellow because there are more oh⁻ than h⁺ ions, and the solution is basic.

the indicator will turn bright red because there are more h⁺ ions than na⁺ ions dissolved in water, and the solution is acidic.

the indicator will not turn colors because there is an equal amount of na⁺ and oh⁻ ions dissolved in water, and the solution is neutral.

Explanation:

1. Analyze NaOH solution: NaOH is a base, so in its aqueous solution, $\ce{OH^-}$ ions are more than $\ce{H^+}$ ions, and the solution is basic.
2. Analyze radish extract indicator behavior: It turns mustard - yellow in high pH (basic) solutions.
3. Evaluate each option:

• Option 1: NaOH solution is basic, not acidic, and $\ce{OH^-}$ ions are more than $\ce{H^+}$ ions, so this is wrong.
• Option 2: NaOH solution has more $\ce{OH^-}$ than $\ce{H^+}$ (basic), and the indicator turns mustard - yellow in high pH (basic) solutions, so this is correct.
• Option 3: NaOH solution is basic, not acidic, and the comparison of $\ce{H^+}$ and $\ce{Na^+}$ is irrelevant for pH - related color change, so this is wrong.
• Option 4: NaOH solution is basic (not neutral), and the ion comparison (equal $\ce{Na^+}$ and $\ce{OH^-}$) is wrong as $\ce{NaOH}$ dissociates into $\ce{Na^+}$ and $\ce{OH^-}$, and the solution is basic, so this is wrong.

Answer:

B. The indicator will turn mustard - yellow because there are more $\ce{OH^-}$ than $\ce{H^+}$ ions, and the solution is basic.