Question

Gout-a condition that results in joint swelling and pain-is caused by the formation of sodium urate (NaC5H3N4) crystals within tendons, cartilage, and ligaments. Sodium urate will precipitate out of blood plasma when uric acid levels become abnormally high. This could happen as a result of eating too many rich foods and consuming too much alcohol, which is why gout is sometimes referred to as the "disease of kings." If the sodium concentration in blood plasma is 0.140 mol L^−1, and Ksp for sodium urate is 5.76 × 10^−8, what minimum concentration of urate would result in the precipitation?

Answer 1

Answer:

The minimum concentration of urate would result in the precipitation is $4.114\times 10^{-7} mol/L$ .

Explanation:

Concentration of sodium ion in blood plasma = $0.140 mol/L$

Concentration of urate ion in blood plasma = $[(C_5H_3N_4)^-]$

The solubility product of sodium urate = $K_{sp}=5.76\times 10^{-8}$

$NaC_5H_3N_4\rightleftharpoons Na^++(C_5H_3N_4)^-$

The expression of solubility product can be given as:

$K_{sp}=[Na^+][(C_5H_3N_4)^-]$

$K_{sp}=0.140 mol/L\times [(C_5H_3N_4)^-]$

$5.76\times 10^{-8}=[(C_5H_3N_4)^-]$

$[(C_5H_3N_4)^-]=4.114\times 10^{-7} mol/L$

If the concentration of an urate ion in blood plasma increases more form $4.114\times 10^{-7} mol/L$ then the precipitation of sodium urate will take place.