Question

Hi!! I need help with no.2 question (in the bottom) please. I am giving you supporting info for your convenience

Purple Phenolphthalein + H+ > Clear Phenolphthalein

Answer 1

NaOH is a strong base(SB) and HCl is a strong acid(SA) . So when SA and SB is titrated and SA is taken in the conical flask, there will be only H+ of HCl in the medium. So the medium is in acidic condition. Purple phenophthalein(PP) is an indicator which we can use in SA-SB titration. It is colorless in acidic medium and purple color in basic medium. When the medium first in acidic condition, as the reaction shows,

Purple phenophthalein + H+ $\leftrightharpoons$ Clear Phenophthalein

the PP is used by the H+ ions and gives a colorless solution which is the clear phenophthalein(CR).

When NaOH is added, the H+ is taken up by the OH- ions of NaOH and nutrelize to H2O, ie,

NaOH + HCl $\leftrightharpoons$ NaCl + H2O

So the PP remains the same since there is no ions like H+ or OH- to react with.

We continue to add NaOH and now all HCl is over, So there is no H+ of HCl in the medium to nutrelize with OH- of NaOH. Then the excess OH- in the NaOH starts to react with PP, ie,

CP + OH- $\leftrightharpoons$ PP (purple color) (basic medium)

So the pH increases and %PP increases. And an equilibrium is established.

Then as we are adding HCl to the medium, the remaining OH- in the medium nutrelize with H+ ions of added HCl, until the all OH- is over. Now the medium is in acidic condition and so the excess H+ ions will react with PP and thus the equilibrium once established is reestablished. The color changes to colorless which is CP.