Item 6 The Henderson-Hasselbalch equation relates the pH of a buffer solution to
ID: 929858 • Letter: I
Question
Item 6
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
You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O.
Express your answer to three significant digits with the appropriate units.
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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 7.40.
Physicians use the following modified form of the Henderson-Hasselbalch equation to track changes in blood pH:
pH=pKa+log[HCO3](0.030)(PCO2)
Part B
If the normal physiological concentration of HCO3 is 24 mM, what is the pH of blood if PCO2 drops to 25.0 mmHg ?
Express your answer numerically using two decimal places.
Item 6
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 a potassium dihydrogen phosphate buffer solution of pH = 6.86. The pKa of H2PO4 is 7.21.You have the following supplies: 2.00 L of 1.00 M KH2PO4 stock solution, 1.50 L of 1.00 M K2HPO4 stock solution, and a carboy of pure distilled H2O.
How much 1.00 M KH2PO4 will you need to make this solution? (Assume additive volumes.)Express your answer to three significant digits with the appropriate units.
Volume of KH2PO4 needed =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 7.40.
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 25.0 mmHg ?
Express your answer numerically using two decimal places.
Explanation / Answer
pH = pka + log [HPO42-]/[H2PO4-]
6.86 = 7.21 + log [HPO42-]/[H2PO4-]
[HPO42-] = 0.4467 [H2PO4-]
moles of HPO42- = 0.4467 x moles of H2PO4- ............(1)
given buffer volume i.e vol of HPO42- + volume of H2PO4- = 250 ml = 0.25 L
let HPO42- volume be V then H2PO4- volume = 0.25-V
now molesof HPO42- = M x V = 1 x V
moles of H2O4- = Mx V = 1 x ( 0.25-V) = 0.25-V
we substitute above in eq (10
V = (0.4467) ( 0.25-V)
V = 0.11167 -0.4467V
V = 0.0772 = 77.2 ml is volume of HPO42- i.e K2HPO4
hence volume of KH2PO4 = 250 -77.2 = 172.8 ml
B) pH = pka + log [HCO3-] ( 0.03pCO2)
pH = 6.1 + log ( 24 x 0.03 x 25)
= 7.36
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