Question

Also use 0.1 M NaOH.

1. The concentration (molarity) of H3PO4 in your 250.00 mL sample using the data from part B.

2. The mass (in mg) of NaH2PO4 in the 250.00 mL sample using the data in part B.

In my data: the first equivalence point is rech at 18 ml at pH 4.75, second equivalence point is reach at 47.0 mL at pH 9.85. I stop titrating at 52 mL with pH 11.00.

Help Please.

Answer 1

ok. First of all you need to find the reaction of dissociation of de H₃PO₄.

H₃PO₄  + H₂O ↔ H₂PO₄^- + H₃O⁺   pKa₁ = 2.1

H₂PO₄^- + H₂O ↔ HPO₄^2- + H₃O⁺ pKa₂ = 7.1

  HPO₄^2- + H₂O ↔ PO₄^3-  + H₃O⁺   pKa₃ =12.3

First point of equivalence=

pH = 4.75 pH = -log [H₃O⁺]   then [H₃O⁺] = 10^-pH

[H₃O⁺] = 1.77x10^-5 M

1) We have the equation [H₃O⁺] = Ka[HA]/[A^-] from the pKa we calculate the Ka with the 10^-pKa

we now from the dissociation that [H₃O⁺] = [A^-]

Ka = 7.9x10^-3

Then the [HA] = [H₃O⁺][A^-]/Ka [HA] = 3.8 x10 ^-8 M = [H₃PO₄]

2) with the same equation but in the second dissociation we calculate the [H₂PO₄^-]

calculate de Ka₂ and the [H₃O⁺] from the pH of the second point.

[H₃O⁺] = 1.41 x10^-10   Ka₂= 7.9x10^-8

now   [HA] = 2.51x10^-13 M = [NaH₂PO₄]

With the volume of the solution we calculate the moles

n = MxV = (2.51x10^-13) (0.25 L) = 6.27x10¹⁴ mol

now with thw MW calculate the g

m= nxMW = 7.46x10^-12g

then we multiplicate for 1000 to find the mg

7.46 x10^-9 mg of NaHPO₄

Hope this works for you!