As I see it you have this:
KIn = 1E-4 = (H^+)(In^-)/(HIn) or
1E-4/(H^+) = (In^-)/(HIn) = 1E3 which tells me that mos of the indicator is in the form In^-. So 2 should be true? What does that tell you about 1?
The indicator HIn has a dissociation constant of 1*10^-4 M.
HIn<===> H30^+ + In^-
Yellow Red
When the pH value of the solution is 7,
1)the indicator shows the colour red
2)at this time most of the indicator has been dissociated
Which statements are true?
1)I took this as false cause the pH range of the indicator is given by pkIn+/-1,so as it indicates the colour red when H3O^+ is released.
So here it should indicate colour yellow.am I correct?
And what about the 2nd one?
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So the solution is acidic? So it should indicate the acidic colour? Am I correct?
Couldn't change the name.My sister was also using the PC to access this site
If most of the indicator is in the In^- form, is that the acid or the base form?
A H atom is released from HIn so it is acidic?
The ratio of In^-/HIn is what counts. The pH tells you what that ratio is, as I showed in the answer to part 2. If most of the indicator is in the acid form it will be yellow; if most in the base form it will be red. The fact that the ionization of HIn produces H^+ doesn't answer the question. The pH of the solution tells the HIn ==> H^+ + In^- equilibrium which way to shift; i.e., to the left as H^+ is increased and to the right as H^+ is decreased.
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