we know that
pka = -log Ka
given
Ka = 1.6 x 10-7
so
pKa = -log 1.6 x 10-7
pKa = 6.7958
so the Pka of the indicator is 6.7958
2)
Let the indicator be HIn
now
HIn ----> H+ + In-
we know that for a indicator
pH = pKa + log [In- / HIn]
given
pH = 9
so
9 = 6.7958 + log [In- / HIn]
[In- / HIn] = 160
as
[In-] >>>> [HIn]
so
the color will be of In- ( ionized form )
so the color of the solution at pH =9 is yellow
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