Question

Buffers in Medicine The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of its conjugate acid and the ratio of the concentrations of the conjugate base and acid. The equation is important in laboratory work that makes use of buffered solutions, in industrial processes where pH needs to be controlled, and in medicine, where understanding the Henderson-Hasselbalch equation is critical for the control of blood pH.

Part A As a technician in a large pharmaceutical research firm, you need to produce 250.mL of 1.00 M a phosphate buffer solution of pH = 7.18. The pKa of H2PO4? is 7.21. You have 2.00 L of 1.00 M KH2PO4 solution and 1.50 L of 1.00 M K2HPO4 solution, as well as a carboy of pure distilled H2O. How much 1.00 M KH2PO4 will you need to make this solution? Express your answer numerically in milliliters to three significant figures. Volume of KH2PO4 needed =   mL   SubmitHintsMy AnswersGive UpReview Part The Henderson-Hasselbalch equation in medicine Carbon dioxide (CO2) and bicarbonate (HCO3?) concentrations in the bloodstream are physiologically controlled to keep blood pH constant at a normal value of 7.4. Physicians use the following modified form of the Henderson-Hasselbalch equation to track changes in blood pH: pH=pKa+log[HCO3?](0.030)(PCO2) where [HCO3?] is given in millimoles/liter and the arterial blood partial pressure of CO2 is given in mmHg. The pKa of carbonic acid is 6.1. Hyperventilation causes a physiological state in which the concentration of CO2 in the bloodstream drops. The drop in the partial pressure of CO2 constricts arteries and reduces blood flow to the brain, causing dizziness or even fainting. Part B If the normal physiological concentration of HCO3? is 24 mM, what is the pH of blood if PCO2 drops to 35.0mmHg ? Express your answer numerically using one decimal place. pH = SubmitHintsMy AnswersGive UpReview Part

Answer 1

The buffer action in phosphate buffer is due to contineous inter conversion of the acid  H2PO₄^- (aq) to the salt HPO₄^2- (aq). .

H2PO₄^-(aq) + H2O(l) ------> HPO₄^2- (aq) + H3O⁺  , pKa = 7.21

Here the required pH = 7.18

Applying Hendersen equation

pH = pKa + log [HPO₄^2-(aq)] / [H2PO₄^-(aq)]

=> 7.18 = 7.21 +  log [HPO₄^2-(aq)] / [H2PO₄^-(aq)]

= > log [HPO₄^2-(aq)] / [H2PO₄^-(aq)] = 7.18 – 7.21 = - 0.03

= > [HPO₄^2-(aq)] / [H2PO₄^-(aq)] / [H2PO₄^-(aq)] = 10^-0.03 = 0.

= > moles of HPO₄^2-(aq) / moles of H2PO₄^-(aq) = 0.9332 ----(1)

Given the volume and concentration of the buffer to be prepared is 250 mL of 1.0M

Hence moles of HPO₄^2-(aq) and H2PO₄^-(aq) in the buffer solution = MxV = 1.0 M x 0.250 L

= 0.250 mol

Hence moles of HPO₄^2-(aq) + moles of H2PO₄^-(aq) = 0.250 mol ---- (2)

From equation (1) and (2) we get

1.9332x moles of H2PO₄^-(aq) = 0.250 mol

=> moles of H2PO₄^-(aq) = 0.250 mol / 1.9332 = 0.129 mol

The available source of H2PO₄^-(aq) is 1.00 M KH2PO₄

Hence volume of 1.00 M KH2PO4 required = 0.129 mol x (1L / 1mol) = 0.129 L (answer)