# Titration of H3PO4 and H2SO4 with methyl orange and phenolphtalein as indicators

There are some problems I don't understand how they are solved (I have the answer). Since I can't find anything related to these on the internet, I'm asking about the problem and the answer here.

### Problem:

For the titration of $$\mathrm{25~mL}$$ solution which has $$\ce{H3PO4}$$ and $$\ce{H2SO4}$$ in it, there are spent $$\mathrm{37.2~mL~0.1008~M~NaOH}$$ in the presence of methyl orange as an indicator. The other $$\mathrm{25~mL}$$ of the solution spent $$\mathrm{49.6~mL~NaOH}$$ in the presence of phenolphthalein as an indicator. Find the concentration of both acids.

In the presence of methyl orange, will happen these reactions:

• $$\ce{H2SO4 + 2 NaOH -> Na2SO4 + 2H2O}$$
• $$\ce{H3PO4 + NaOH -> NaH2PO4 + H2O}$$

And there are spent $$\mathrm{37.2~mL~NaOH}$$.

In the presence of phenolphthalein, the following reactions will happen:

• $$\ce{H2SO4 + 2NaOH -> Na2SO4 + 2H2O}$$
• $$\ce{H3PO4 + 2NaOH -> Na2HPO4 + 2H2O}$$

And there are spent $$\mathrm{49.6~mL~NaOH}$$.

### Question:

The procedure to find the concentration of the acids is obviously easy, but what I don't understand is the volumes of $$\ce{NaOH}$$ we should take to make the calculations.
The volume of $$\ce{NaOH}$$ needed for $$\ce{H3PO4}$$ is $$\mathrm{49.6-37.2=12.4~mL}$$.
The volume of $$\ce{NaOH}$$ in presence of methyl orange, needed for $$\ce{H2SO4}$$ is $$\mathrm{37.2-12.4=24.8~mL}$$, and in the presence of phenolphthalein is $$\mathrm{49.6-24.8=24.8~mL}$$.

There are also more similar problems with different compounds, but I totally don't understand the logic how these volumes are calculated! How do you know which one to deduct?
If you also know any website with these problems please send me the link, thanks.

• Subtracting volumes, while 100% legitimate, is a peculiar and unobvious thing to do. Personally, I'd rather go with molar quantities. Let's say your 25 mL sample contains a moles of $\mathrm{H_3PO_4}$ and b moles of $\mathrm{H_2SO_4}$. How many moles of NaOH you would then need for the first titration? What about the second? Sep 3 '15 at 8:16
• You say that it is obviously easy and that you have the answer. But you don't show the obviously easy way and you don't tell about the answer. Could you please add those things? Sep 3 '15 at 9:15
• The formula for the calculation is C (H2SO4) x V (H2SO4) = C (NaOH) x V (NaOH) Sep 3 '15 at 9:17
• For all its worth, here are the answers : C(H2SO4)= 0.05 M and C(H3PO4)=0.05 M Sep 3 '15 at 9:19

I for myself like to watch titration curves, because they often give hints about where to look. So the figures below show you the theoretical titration curves of your both acids$^{\ast 1,2}$ each and also titrated together with an aqueous solution of NaOH$^{\ast 3}$.

But ... how exactly can this be helpful?

1. You can see that at the "first" inflection point, which methyl orange would detect, both acids lost one to two protons. So how many equivalents of $\ce{NaoH}$ was needed?
2. At the "second" inflection point, which is detected by phenolphthalein, only phosphoric acid has another inflection point. And how many equivalents of $\ce{NaoH}$ was needed now?

With those thoughts in mind, the question can also be answered without my fancy titration curves.

You need to solve this equation system to get both concentrations: \begin{align}3 c_B V_B &= 2 c_{S1} V_i+c_{S2} V_i\\4 c_B V_B &= 2 c_{S1} V_i+2 c_{S2} V_i\end{align} where $V_i$ is the initial volume of both acids ($\mathrm{25~mL}$).

$^{\ast 1}$ $\mathrm{pK_a}$ of $\ce{H2SO4}$: $-3,~1.9$ (according to the german wikipedia)

$^{\ast 2}$ $\mathrm{pK_a}$ of $\ce{H3PO4}$: $2.16,~7.21,~12.32$ (according to the german wikipedia)

$^{\ast 3}$ the increasing volume is neglected and both acids are assumend to have the same concentration

• Thanks, but i won't have time for curves in the exam ! Sep 4 '15 at 7:31
• As I said, you don't need the curves. Sep 7 '15 at 12:04