You have titrated 10.00 mL sample solution of mixed acid solution (HCl and H3PO4) with 0.09982 M NaOH, which required 19.47 mL of NaOH to reach

Question

You have titrated 10.00 mL sample solution of mixed acid solution (HCl and H3PO4) with 0.09982 M NaOH, which required 19.47 mL of NaOH to reach the first equivalence point and 33.51 mL to reach the second equivalence point.

1. Calculate the phosphoric acid concentration, [H3PO4], in the mixed acid solution.

2. Calculate the concentration of the HCl in the mixture in #2 above. Recall that the first equivalence point represents the sum of the protons from HCL dissociation and the first protons from phosphoric acid dissociation.

Answer 1

Finding the concentration of H₃PO₄ requires us to separate it from the HCl. One nice thing is that we are given the second equivalence point, something that HCl does NOT contribute to, but H₃PO₄ does. More specifically, the second equivalence point refers to the solution's ability to continue to donate protons as H₂PO₄^-.

Therefore, the difference between the second equivalence point and the first equivalent point tells us only about the amount of H₃PO₄ in the original acid solution.

33.51 mL - 19.47 mL = 14.04 mL NaOH required to get from the first equivalence point to the second one.

14.04 mL NaOH * (1 L / 1000 mL) * 0.09982 (mol/L) = 0.001401 moles of NaOH to get from 1st equiv point to 2nd equiv point.

This means that there must've been 0.001401 moles of H₃PO_4, because each mole of NaOH can neautralize one H₂PO₄^-. The concentration is therefore 0.001401 mol / 10 mL * (1000 mL / L) = 0.1401 mol/L, or [H₃PO₄] = 0.1401 M

We can now find the concentration of HCl by using the first equivalence point.

19.47 mL NaOH required to neutralize both HCl and H₃PO₄ to H₂PO₄^-.

19.47 mL * (1 L / 1000 mL) * 0.09982 (mol/L) = 0.001943 moles of NaOH

Of the 0.001943 moles of NaOH that we used to reach the first equiv point, 0.001401 moles (value calculated from before) were used to neutralize H₃PO₄, while all the rest were used for HCl.

0.001943 - 0.001401 = 0.0005425 moles HCl

0.0005425 mol / 10 mL * (1000 mL / L) = 0.05425 mol/L, or [HCl] = 0.05425 M