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The solution can contain $\ce{HCl}$, $\ce{H3PO4}$ or $\ce{NaH2PO4}$ individually or in all possible combinations. An aliquot of $25.00\,\text{mL}$ of this solution consumes $14.56\,\text{mL}$ of $0.1000\,\text{M}$ $\ce{NaOH}$ solution with methyl orange, while another aliquot of $25.00\,\text{mL}$ of the same solution consumes $47.50\,\text{mL}$ of base with phenolphthalein. Determine the composition of the solution.

Much less base was used with methyl orange than with phenolphthalein. Since methyl orange changes color in the acidic region, then the stated volume of base is used for either $\ce{HCl}$ or $\ce{H3PO4}$ or both. How to determine this exactly?

In addition to phenolphthalein, a substance (or substances) was proven which was also proven with methyl orange and some additional substance. However, I don't understand how to determine the exact composition.

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  • $\begingroup$ Hint: Evaluate the ratio of spent volumes. $\endgroup$
    – Poutnik
    Jan 9 at 16:56
  • $\begingroup$ Hmm, the ratio is around 0.3, but I don’t know how to use that further. $\endgroup$ Jan 9 at 18:36
  • $\begingroup$ If V2/V1 = 2, then there is H3PO4. If V2/V1 < 2 then there is HCl + H3PO4. If V2/V1 > 2 There is H3PO4 + NaH2PO4. If there was initially added HCl + NaH2PO4, than it depends what of these two gets the upper hand, forming H3PO4. $\endgroup$
    – Poutnik
    Jan 9 at 19:15
  • $\begingroup$ I’m sorry, but could you explain a bit more? How did you get to that? $\endgroup$ Jan 9 at 19:55

1 Answer 1

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It is difficult, but not impossible to solve this problem, namely find $3$ concentrations out of two measurements. Let's try.

The two titrations give two useful volumes of $\ce{NaOH}$ $0.1$M.: $14.56$ mL and $47.50$ mL, which correspond to amount of $\ce{NaOH}$ equal to $1.456$ mmol and $4.75$ mmol. The acidic samples to be titrated contain the same total amounts of $\ce{H+}$ in $25$ mL. Their acidic concentrations $\ce{[H+]}$ are equal to $1.456$ mmol/$25$ mL = $0.0582$ mol/L in the first titration and $4.75$ mmol/$25$ mL = $0.19$ mol/L in the 2nd titration.

The first titration is done with methylorange, that is up to pH $4$, where $\ce{HCl}$ and $\ce{H3PO4}$ are titrated.

The second titration is done with phenolphtalein, that is up to pH $9$, where $\ce{HCl, H3PO4 twice}$ and $\ce{H2PO4^-}$ from $\ce{NaH2PO4}$ are titrated.

$$\ce{[HCl] + [H3PO4]} = 0.0582 M \tag{1} $$

$$\ce{[HCl] + 2[H3PO4] + [NaH2PO4]} = 0.19 M \tag{2}$$

If {$2$} is subtracted from ($1$) it gives : $$\ce{[H3PO4] + [NaH2PO4] = 0.1318} M \tag{3} $$

But the data says that the solution contains either $\ce{H3PO4}$ or $\ce{NaH2PO4}$. Both products never occur simultaneously. If $\ce{H3PO4}$ exist in solution, its concentration must be < $0.058$ M (from ($1$), and $\ce{[NaH2PO4]}$ must be higher than $0$, from ($3$), (must be between $0$ and about $0.1..$ M ). This is contrary to the initial text : only one of the two products $\ce{H3PO4}$ and $\ce{NaH2PO4}$ must be present. We just saw that if $\ce{H3PO4}$ is present, $\ce{NaH2PO4}$ must also be present in solution. This is excluded. So $\ce{H3PO4}$ is not in solution. As a consequence : $$\ce{[H3PO4]} = 0$$ and $\ce{NaH2PO4}$ is the only phosphorus containing compound in solution. The final composition of the acidic solution is : $$\ce{[HCl] = 0.0582 M}$$ $$\ce{[NaH2PO4] = 0.1318 M} $$

This is all the job to be done.

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  • $\begingroup$ I think you misinterpreted the text. In my opinion, it does not mean HCl AND ( H3PO4 XOR NaH2PO4). Chemically, If there is no H3PO4, there is no HCl either. $\endgroup$
    – Poutnik
    Jan 9 at 17:51
  • $\begingroup$ @Poutnik. Try to solve the problem without my understanding. You will fail $\endgroup$
    – Maurice
    Jan 9 at 20:59

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