Question

Uric acid can collect in joints, giving rise to a medical condition known as gout. If the pKa of uric acid is 3.89, what is the pH of a 0.0140 M solution of uric acid?

Answer 1

Answer:

pH = 2.87

Explanation:

Let us represent uric acid as HA

The dissociation will be:

                     $HA ---> H^{+} + A^{-}$

Let "x" concentration of HA dissociates so it will give x concentration of hydrogen ion and the conjugate base.

$Ka=\frac{[H^{+}][A^{-}]}{[HA]}$

pKa = 3.89

Therefore Ka = 0.000129

Putting values

$0.000129 = \frac{x^{2} }{(0.014-x)}$

on may ignore x in denominator

x = 0.00134 M

[H⁺] = 0.00134

pH = -log[H⁺] = 2.87

Answer 2

The pH of the solution is 2.89.

Let the solution of the uric acid be HA. We have to set up the ICE table as follows;

         HA(aq) + H20(l)  ⇄ H3O^+(aq)  + A^-

I         0.0140                       0                 0

C       -x                                +x                +x

E    0.0140 - x                      x                   x

Given that;

Ka = [ H3O^+]  [A^-]/HA

Ka = Antilog (-Ka)

Ka = Antilog (-3.89) = 1.3 × 10^-4

1.3 × 10^-4 = x^2/0.0140 - x  

1.3 × 10^-4(0.0140 - x) = x^2

1.82 × 10^-6 - 1.3 × 10^-4x  = x^2

x^2 + 1.3 × 10^-4x - 1.82 × 10^-6 = 0

x = 0.0013M

Hence;

[ H3O^+] =  [A^-] = 0.0013M

pH = -log(0.0013M)

pH = 2.89

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