Respuesta :
In order to make 100 mL of the pH 7.00 buffer, the quantities of the equimolar aqueous solutions of K₂HPO₄=447ml and KH₂PO₄ = 353ml must be combined.
Steps
In this instance, two salts, K2HPO4 and KH2PO4, are present. These salts will dissociate to produce the ions HPO42 and H2PO4, which are derived from the acid H3PO4, a polyprotic acid. There are 3 pKa values for this acid:
H₃PO₄ ⇄ H⁺ + H₂PO₄⁻ (pKa1 = 2.15)
H₂PO₄⁻ ⇄ H⁺ + HPO₄²⁻ (pKa2 = 7.10)
HPO₄²⁻ ⇄ H⁺ + HPO₄³⁻ (pKa3 = 12.40)
H2PO4 is the acid and HPO42 is the conjugate base in this equilibrium, which is the second. Thus,
7.00 = 7.10 + log[HPO₄²⁻]/[H₂PO₄⁻]
log[HPO₄²⁻]/[H₂PO₄⁻] = -0.10
[HPO₄²⁻]/[H₂PO₄⁻] =
[HPO₄²⁻]/[H₂PO₄⁻] = 0.79
The ions have the same concentration as the salts. When they are joined in an equimolar solution, they both have the same number of moles. The concentration is calculated by dividing the volume by the number of moles. Using K2HPO4 as 1 and KH2PO4 as 2 as a result:
(n1/V1)/(n2/V2) = 0.79
(n1/V1)*(V2/n2) = 0.79
Because n1 = n2,
V2/V1 = 0.79
V2 = 0.79V1
There must be 800.0 mL in total, therefore
V1 + V2 = 800.0
V1 + 0.79V1 = 800.0
1.79V1 = 800.0
V1 = 447.0 mL
V2 = 353.0 mL
The equimolar aqueous solutions of K₂HPO₄=447ml and KH₂PO₄ = 353ml must be combined to make 100 mL of the pH 7.00 buffer.
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