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What is the pKa of a compound used to prepare a buffer of pH 10.65 using...

What is the pKa of a compound used to prepare a buffer of pH 10.65 using 3 times as much conjugate base than the weak acid?

The solubility of sodium sulphate (Na2SO4) in water is 9.4 g/100 mL. Calculate the Ksp for sodium sulphate.

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Answer #1

ratio of conjugate base / acid = 3/1

pH = pKa + log [conjugate base / acid]

10.65 = pKa + log [3]

pKa = 10.17

b)

solubility S = 9.4 g / 100 mL

moles = 9.4 / 142 = 0.0662

Molarity = 0.0662 / 0.1 = 0.662 M

Na2SO4 -------------------> 2 Na+ + SO42-

                                      2 S            S

Ksp = [Na+]^2[SO42-]

       = (2S)^2 x S

       = 4 S^3

       = 4 x 0.662^3

       = 1.16

Ksp of Na2SO4 = 1.16

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