If the sodium concentration in blood plasma is 0.140 M, and K_sp for sodium urate is 5.76*10^-8, what minimum concentration of urate would result in the precipitation of odium urate?
Dr.A
The equilibrium
C5H3N4O3Na <-------> C5H3N4O3- + Na+
requires that
Ksp = [C5H3N4O3-][Na+]
let x = mol/L sodium urate that dissolve : this will produce x mol/L urate and x mol/L Na+.
But there is arleady some Na+ so at equilibrium
[Na+]= 0.140 +x and [urate] = x
5.76 x 10^-8 = (0.140+x)(x)
since x has to be small compared to 0.140 we can assume 0.140+x = x.
this gives us the approximate equation
5.76 x 10^-8 = (0.140)(x)
x = [urate]=4.11 x 10^-7 M
Orignal From: Concentration/Equilibrium Math Problem HELP PLEASE?
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