Consider an indicator ionized as shown. for which K H In = 1.0 x 10 -4 yellow/re
ID: 910763 • Letter: C
Question
Consider an indicator ionized as shown.
for which KHIn = 1.0 x 10-4 yellow/red and consider the statements
A) the predominant color in its acid range is yellow.
B) In the middle of the pH range of its color change a solution containing the indicator probably will be orange.
C) at PH=7 a solution containing this indicator (and no other colored spiecies) will be red.
D) at PH= 7 most of the indicator is not ionized.
Which of the responses contains all the true statements and no others? Write the equilibrium constant expression for the indicator.
Answers
1. another combination
2. c and d
3. b and d
4 a, b, and c
5 a and c
Explanation / Answer
Hln + H2O = H3O+ + In-
Yellow red
KHln = [H+][ln-]/[Hln] = 10-4
To the half-way through the colour change, the concentrations of the acid and its ion are equal. In that case, they will cancel out of the KHln expression.
At this point, [H+] = Khln = 10-4
pH = 4 ....(This is the middle of the pH range of its color change)
A. The predominant color in its acid range(Hln) is yellow.
B. At the middle point of pH range (at pH=4), the concentrations of the two colours(yellow and red) will become equal. The colour you see will be a mixture of the two which is yellow.
C. at PH=7 (>4) a solution containing this indicator will be predominant in ln- , so the colour will be red.
D. At pH= 7 (>4) the indicator will be almost completely ionized.
So answer is 4 ( A, B , C are correct)