A 125 mL sample of phosphoric acid was titrated with an average of 63.7 mL of a 0.050M barium hydroxide solution according to the given reac

Question

A 125 mL sample of phosphoric acid was titrated with an average of 63.7 mL of a 0.050M barium hydroxide solution according to the given reaction: 2 H3PO4 + 3 Ba(OH)2 ⇌ Ba3(PO4)2 + 6 H2O Determine the concentration of the phosphoric acid sample:

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Diễm Thu 6 months 2021-07-15T02:50:33+00:00 1 Answers 0 views 0

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    2021-07-15T02:52:23+00:00

    Answer:

    0.017 M

    Explanation:

    The balanced equation for the reaction is given below:

    2H₃PO₄ + 3Ba(OH)₂ ⇌ Ba₃(PO₄)₂ + 6H₂O

    From the balanced equation above,

    The mole ratio of acid, H₃PO₄ (nₐ) = 2

    The mole ratio of base, Ba(OH)₂ (n₆) = 3

    Finally, we shall determine the concentration of the phosphoric acid, H₃PO₄. This can be obtained as follow:

    Volume of acid, H₃PO₄ (Vₐ) = 125 mL

    Volume of base, Ba(OH)₂ (V₆) = 63.7 mL

    Concentration of base, Ba(OH)₂ (C₆) = 0.050M

    Concentration of acid, H₃PO₄ (Cₐ) =?

    CₐVₐ / C₆V₆ = nₐ/n₆

    Cₐ × 125 / 0.05 × 63.7 = 2/3

    Cₐ × 125 / 3.185 = 2/3

    Cross multiply

    Cₐ × 125 × 3 = 3.185 × 2

    Cₐ × 375 = 6.37

    Divide both side by 375

    Cₐ = 6.37 / 375

    Cₐ = 0.017 M

    Thus, the concentration of the phosphoric acid, H₃PO₄ is 0.017 M

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