Question

If the sodium concentration in blood plasma is 0.140 M, and Ksp for sodium urate is 5.76×10−8, what minimum concentration of urate would result in the precipitation of odium urate?

Answer 1

The equilibrium

C5H3N4O3Na ---------> C5H3N4O3^-   +Na⁺

Ksp    = [C5H3N4O3-] [Na⁺]
let x moles/L sodium urate dissolve will produce x mole/L urate and x moles of Na⁺
but already some Na+ present at equilibrium
[Na+] = 0.140+x and [Urate] =x

Ksp    = [C5H3N4O3-] [Na⁺]

5.76*10^-8    = x*(0140+x)

5.76*10^-8 = 0.14x + x²

x² + 0.14x-5.76*10^-8 =0

     x= 4.1*10^-7 M

[urate] = x = 4.1*10^-7 M