Question

A phosphate buffer is involved in the formation of urine. The developing urine contains H2PO4 and HPO42- in the same concentration as present in blood plasma. Identify the acid and its conjugate base. Write the ionization equation and the Ka expression. 6.2 X 10^(8) a. The Ka of H2PO4 is 6.2 x 108. What is the pka? Is this acid weaker or stronger than H2CO3? Which buffer system is more optimal for regulating pH in the body? b. Using LeChatlier's Principle, what happens if the urine is acidified (H+ ions added)?

Answer 1

The ionization equation is

`H_(2)PO_(4)^(-)  +H_(2)O` ⇄
`HPO_(4)^(-2)  +H_(3)O^(+)` (1)

Step-by-step explanation:

The ionization equation is

`H_(2)PO_(4)^(-)  +H_(2)O` ⇄
`HPO_(4)^(-2)  +H_(3)O^(+)` (1)

As the Bronsted definition sais, an acid is a substance with the ability to give protons thus, H2PO4 is the acid and HPO42- is the conjugate base.

The Ka expression is the ratio between the concentration of products and reactants of the equilibrium reaction so,

`Ka = ([HPO_(4)^(-2)] [H_(3)O^(+)])/([H_(2)PO_(4)^(-)] [H_(2)O]) = 6.2x10^(-8)`

The pKa is

`-Log (Ka) = -Log (6.2x10^(-8)) = 7.2`

The pKa of H2CO3 is 6,35, thus this a stronger acid than H2PO4. The higher the pKa of an acid greater the capacity to donate protons.

In the body H2CO3 is a more optimal buffer for regulating pH due to the combination of the two acid-base equilibriums and the two pKa.

If the urine is acidified, according to Le Chatlier's Principle the equilibrium (1) moves to the left neutralizing the excess proton concentration.