Kidneys are the body\'s natural defense mechanism against acidosis. The proximal

Question

Kidneys are the body's natural defense mechanism against acidosis. The proximal
tubules produce NH3 to counter the effects of hydrogen ions by removing them from
the blood stream to form NH4+. Assuming that the blood pH is at 7.2, how many of
moles of NH3 must be produced to raise the blood pH back to a normal 7.4? The Ka
of NH4+ is 5.6 x 10^-10 M. You can assume that there is no NH3 initially in the
blood. The initial concentration of NH4+ in the blood is 60 ug/100 mL. Assume
that the bicarbonate buffering system is at normal levels for a blood pH of 7.2.

Please provide help using either matlab or mathcad.

Thanks in advance.

Explanation / Answer

NH3 + H20 ------> NH4+ + OH-

its the reactions taking place there

given the Ka of NH4 + is 5.6 x 10^-10 M.

so Kb of ammonia = Kw / Ka = 1.78 * 10^-5 [Ka = 10^-14]

here NH3 and NH4 + forms a buffer system

by buffer formula ;

pH = pKb + log [NH4+] / [NH3]

initial concentration of NH4 + = 60*10^-6 gm /100ml

per litre =60 *10^-5gms

moles of NH4+ = 3.33*10^-5

concentration in 1 litre = 3.33 * 10^-5 M

pKb = 4.74

keeping in above equation; for pH = 7.2

[NH3] = 9.60 * 10^-3

so moles of NH3 = 9.60 * 10^-3 since we hadd taken 1 litre.

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