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6 March, 01:26

Phosphoric acid is a triprotic acid with the following pKa values:

pKa1=2.148, pKa2=7.198, pKa3=12.375

You wish to prepare 1.000 L of a 0.0500 M phosphate buffer at pH 7.540. To do this, you choose to mix the two salt forms involved in the second ionization, NaH2PO4 and Na2HPO4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture?

a. Mass NaH2PO4

b. Mass Na2HPO4

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  1. 6 March, 01:44
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    Mass NaH₂PO₄ = 1.920 g

    Mass Na₂HPO₄ = 4.827 g

    Explanation:

    For a buffer solution we know its pH can be calculated from the Henderson-Hasselbach formula:

    pH = pKa + log [A⁻]/[HA]

    where [A⁻] and [HA] are the concentrations of the weak acid and its conjugate base in the buffer.

    We want to prepare a buffer at pH 7.540 so we have chosen salts NaH₂PO₄ and Na₂HPO₄ as the weak acid and conjugate base respectively.

    To calculate the mass of these salts we need to compute their ratio in the Henderson - Hasselbach equation.

    Now since we are asked to determine the masses of NaH₂PO₄ and Na₂HPO₄ and we know we want to prepare 1.000 L of a 0.05 M phosphate buffer, we can setup a system of 2 equations with two unknowns from the ratio mentioned above:

    pH = pKa + log [A⁻]/[HA]

    7.540 = 7.198 + log[HPO₄²⁻] / [H₂PO₄ ⁻]

    0.342 = log[HPO₄²⁻] / [H₂PO₄ ⁻]

    taking inverse log function to both sides of this equation:

    2.198 = [HPO₄²⁻] / [H₂PO₄ ⁻]

    but this is also equivalent to

    2.198 = mol HPO₄²⁻ / mol H₂PO₄⁻ (M = mol/V)

    We also know that in 1 liter of 0.05 M phosphate, we have 0.05 total mol HPO₄²⁻ and H₂PO₄⁻, thus

    mol HPO₄²⁻ + mol H₂PO₄⁻ = 0.05 mol

    2.198 = mol HPO₄²⁻ / mol H₂PO₄⁻

    solving this system of equations calling x = mol HPO₄²⁻ and y = mol H₂PO₄⁻, we have:

    2.198 = x / y ⇒ x = 2.198y

    x + y = 0.05

    2.198y + y = 0.05

    3.198 y = 0.05 ⇒ y = 0.05 / 3.198 = 0.016

    x = 0.05 - 0.016 = 0.034

    and the masses can be calculated from the molar masses (141.96 g/mol Na₂HPO₄ and 119.98 g/mol NaH₂PO₄

    mol HPO₄²⁻ = 0.034 mol x 141.96 g/mol = 4.827 g

    mol H₂PO₄⁻ = 0.016 mol x 119.98 g/mol = 1.920 g
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