Question

A patient is 6-feet tall and has a body mass between 190 and 225 lb so their blood volume is approximately 5.8L. The patient's blood pH, due to cardiac arrest is 7.00. How many mL of 84 mg/mL bicarbonate must be administered intravenously to bring the patient's blood pH to between 7.35 and 7.45?

Answer 1

Assuming that we want the pH to be 7.4

So,

pOH = 14-7.4 = 6.6

So,

[OH^-] = 10^-pOH = 10^-6.6 M

For carbonic acid, K_a = 4.5*10^-7

The reaction taking place is:

HCO₃^- + H₂O ---> H₂CO₃ + OH^-

So for this reaction, K_b = 10^-14/K_a

Using Ostwald's dilution law:

[OH^-] = (K_b*C)^0.5

Putting values:

10^-6.6 = ( ((10^-14)/(4.5*10^-7))*C )^0.5

Solving we get:

C = 2.84*10^-6 M

Stock conc of bicarbonate present = 84 mg/ml = 84 g/L = (84/61) mol/L = 1.37 M

So, dilution factor = 1.37/C = 1.37/(2.84*10^-6) = 4.82*10⁵

Assume that 'x' L of this bicarbonate stock is added.

We have:

(x+5.8)/x = 4.82*10⁵

Solving we get:

x = 0.000012 L

So,

Volume of this stock that must be administered = 0.012 mL

Hope this helps !