Question

Hi! I need help on this lab report question and I can't seem to figure it out!

Essentially, in this experiment, we titrated 0.095189 M NaOH into a 0.5 M phosphoric acid solution. The volume of NaOH at the first equivalence point is 3.66 and the corresponding pH is 3.61.

What is the concentration of the phosphoric acid based on the 1^st Equivalence Point? (M)

We're given the volume of Na

Answer 1

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Titratten of H3 POq us. NaOH : en The reaction is Itz PO4 + NaOH > Nalto POA +H2O We have to calculate the concentration of H₂PO4 at the 1st equiralance point. Given i Strength of Naon Strength of NaOH (5.1 = 0.095189 (my strength of H3 PO4 (53] = 0.5 (M) Volume of NaOH aut ist equivalance point (U) = 3.66 ML Volume of H3 POq at ist equivalance point = V2 fay) Now, VI SI = V2 sz W, V2 - YSL = 3.66 ML X 0.095189 (M) V2 = 0.6967 ML ~0.70 ML i Now, total volume of the soutien = 3:66 069 (3.66 +0.70) mL = 4.36 m2 Amount of H3 P04 added to the solution = 0.70 mL X 0.5 j = 0.35 ml x me = 0.35 m-mol Total volume of solution - 4136 mL :. Concentration of H3 PO4 = 0.35 m.mol 4:36 mm i = 0.08027 mol , = 0.08027(My : Concentration of H3 P09 at ist equiualance point-05080