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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?

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  • $\begingroup$ Using $\ce{ insert chemical here }$ will save you time and keystrokes. Not to mention it looks neater. $\endgroup$ – Berry Holmes May 22 '17 at 6:49
  • $\begingroup$ Phenolphthalein changes color only above a $\text{pH}$ of about 11 or 12. So it shows a change only after quite alot of base is added. $\endgroup$ – Pritt Balagopal May 22 '17 at 7:56
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a) That's right, $\ce{NaOH}$ reacts first. But you are wrong in thinking that $\ce{NaCl}$ has a pH of 7. In a way, $\ce{NaCl}$ has "no pH at all". It just sits there and does nothing. It is other component(s) of the solution that determine its pH. At that point, the other component is $\ce{Na2CO3}$, which is alkaline due to hydrolysis. As the titration progresses, it turns to $\ce{NaHCO3}$, hence the change in pH.

b) Well, $\ce{H2CO3}$ would indeed make the solution slightly acidic, but not enough so to trigger the color change in methyl orange. It is $\ce{HCl}$ that does the job. After the neutralization is completed, one tiny extra drop is enough.

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