Let us consider a mixture of $\ce{NaOH}$ and $\ce{Na2CO3}$. Based on what I have understood, double titration works in the below manner :
Phenolpthalein turns pink when added to the mixture. As $\ce{HCl}$ is added, the pH of the mixture keeps reducing and phenolpthalein becomes colorless when the mixture becomes of a pH less than $8.3$ ie. at a stage when $\ce{NaOH}$ is completely neutralised and $\ce{Na2CO3}$ is half neutralized. Then, we add methyl orange which makes the solution yellow. As we add more $\ce{HCl}$, the pH reduces and hence, methyl orange turns red when the neutralization of $\ce{Na2CO3}$ is complete.
My questions are:
a) When phenolpthalein is added, I have read elsewhere that $\ce{NaOH}$ reacts with $\ce{HCl}$ first. In that case, phenolpthalein should make the solution colorless as $\ce{NaOH}$ neutralizes with $\ce{HCl}$ to form $\ce{NaCl}$ which has a pH of $7$ (not between $8.3$ and $10.5$). Why doesn't phenolpthalein change its color until $\ce{Na2CO3}$ is also half neutralized? Is it because the reactions take place simultaneously?
b) When methyl orange is added, $\ce{NaHCO3}$ is neutralized to $\ce{NaCl}$. But pH range of methyl orange is 3-4.5, but the solution obtained is neutral. How does methyl orange change its color, then?I think it is because of the $\ce{H2CO3}$ ($\ce{H2O + CO2}$) that is formed along with $\ce{NaCl}$, which makes the solution slightly acidic. Is that a correct explanation?
$\ce{ insert chemical here }$
will save you time and keystrokes. Not to mention it looks neater. $\endgroup$