When H+ is added to the H2CO3-HCO3^- buffer, CO2 (g) + H2O(l) H^(+) + HCO^ –Then added hydrogen ion is used up by the reaction with HCO3^– ions. When H^+ is removed, the equilibrium shifts to...

When H+ is added to the H2CO3-HCO3^- buffer, CO2 (g) + H2O(l) H^(+) + HCO^ –

Then added hydrogen ion is used up by the reaction with HCO3^– ions. When H^+ is removed, the equilibrium shifts to the right, forming more hydrogen ions. The H2CO3-HCO3^- buffer also buffers saliva, removing hydrogen ion produced by bacteria in the mouth and from acids in foods. This buffering of saliva helps prevent cavities. The H2CO3-HCO3^- buffer cannot buffer against acidity resulting from carbon dioxide. This function is carried out mainly by hemoglobin. A H2PO4^(-)–HPO4^(2-) buffer system

H2PO4^- <->H^(+) + HPO4^(2-)

is important inside cells and in the urine. Urine may also be buffered by an ammonia-ammonium ion buffer

NH3(aq) + H^+ <->NH4^+

For the H2PO4^(-)-HPO4^(2-) buffer system, pKa = 7.21. If the pH of a urine sample is 5.6 and its total phosphate concentration (i.e. [H2PO4^- ] + [HPO4^(2-) ]) is 46.6 mmol/L, calculate the concentration of each phosphate species in the sample.

Asked on by bobbyjimbo

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llltkl | College Teacher | (Level 3) Valedictorian

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The phosphate buffer system involves a single proton transfer, hence it should follow the Henderson equation of the form pH = pKa + Log [HPO4=]/[H2PO4-]. 

Let q be the no. of millimoles of [HPO4=] per litre, then no. of millimoles of [H2PO4-] per litre would be (46.6-q)

By equation, 5.6=7.21+Log[q]/[46.6-q]

or, q =1.11649

i.e. no. of millimoles of [HPO4=] per litre =1.11649

no. of millimoles of [H2PO4-]  per litre = 46.6-q = 45.4835

 

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