- #1
Qube
Gold Member
- 468
- 1
Homework Statement
Indicator HIN is yellow at pH 1.0 or less, green at pH 2.0, and blue at pH 3.0 or above.
A 0.004 M sample of colorless acid HA is treated. It develops a green-blue color halfway between pure green and pure blue.
How do the strengths of HA and HIN compare?
Homework Equations
[itex]K_{a} = \frac{[H_{3}O^{+}][IN^{-}]}{[HIN]} = \frac{[H_{3}O^{+}][A^{-}]}{[HA]}[/itex]
The Attempt at a Solution
Midway between the pH values for indicator HIN, or at pH 2.0, the hydronium ion concentration is equal to the Ka value. The hydronium ion concentration is [itex]10^{-2}[/itex]. The [itex]K_{a}[/itex] value is therefore [itex]10^{-2}[/itex].
The same goes for the acid HA. Since at pH of 2.5, half of the acid has been ionized to hydronium ions, the strength of the acid, or Ka, is equal to [itex]10^{-2.5}[/itex].
Therefore, through comparison of the Ka values, HIN is a stronger acid, as it has the larger Ka value.
1) Is this correct? Is the train of thought also correct? I think I'm missing something here.
Last edited: