Question

I have already posted this questions once... but some doofus responded with "additional information needed". But I posted all the information that the question gives, word for word. So i'm going to give it another shot. Here is a similar question to Part A. (http://www.chegg.com/homework-help/questions-and-answers/technician-large-pharmaceutical-research-firm-need-produce-450ml-100m-potassium-phosphate--q1392759)

± 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 350. mLof a potassium dihydrogen phosphate buffer solution of pH = 6.99. 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.)

Volume of KH2PO4 needed = ? mL

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 CO2is 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 27.0 mmHg ?

pH = ?

Answer 1

A)

let

y L of K2HP04 be taken

then

volume of KH2P04 = 0.35 - y

we know that

moles = molarity x volume (L)

so

moles of K2HP04 = 1 x y = y

moles of KH2P04 = 1 x (0.35-y) = 0.35-y

now

we know that

for buffers

pH = pKa + log [base /acid]

so

pH = pKa + log [ K2HP04 / KH2P04]

6.99 = 7.21 + log [ K2HP04 / KH2P04]

[ K2HP04 / KH2P04] = 0.60256

as the final volume is same for both

moles of K2HP04 / KH2P04 = 0.60256

moles of K2HP04 = 0.60256 x moles of KH2P04

so

y = 0.60256 x ( 0.35 - y)

y = 0.1316

so

volume of K2HP04 = 0.1316 L = 131.6 ml

volume of KH2P04 = 350 - 131.6 = 218.4 ml

B)

given

pH = pKa + log [HC03-] / (0.03 x PC02)

given

[HC03-] = 24 mM

pC02 = 27

pKa = 6.1

so

using those values

we get

pH = 6.1 + log [ 24 / 0.03 x 27]

pH = 7.57

so

pH of the blood is 7.57