ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: tcg93 on October 20, 2010, 11:42:49 am
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HIN is an indication that is a weak acid with Ka = 2.0 x 10^(-5). It is yellow in acidic form and green in basic form. The equilibrium reaction is
HIn <=> H+ + In-
If HIn is added to an acidic solution of PH 1.0 equilibrium shifts to the ___ and the solution turns ____.
A) Right, Green
B) Left, Yellow
C) Right, Yellow
D) Left, Green
answer below
why is the answer B????
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.. my answer was A.....
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HIn <=> H+ + In-
the HIn that is added would itself be at equilibrium, so when mixed with the strong acid(ph1) full of H+ it acts to partially oppose this change thus using up H+ to create more HIn
YET
it's a weak acid and only acts to PARTIALLY oppose the change, as it is such a strong acid that it is added to with such a low PH it would not make much difference, ie. the ph may increase to 2,
and then it'd still be crazy acidic with the ph of <7; so yellow.
though admittedly at first glance i picked C.
... had to re-read and think about this for far too long.
chemistry multiple choice aggravates me.
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my understanding is that first we have an strong acid solution that is at equilibrium. we add HIn to it and the HIn has that equation they write, so in effect you are adding a small amount of H+ ions, now the system will partially oppose the change by shifting equilibrium to the left, we get less H+ as a result but still more H+ them before we added the HIn and therefore the solution is still acidic, giving us a yellow colour