A 70-kg person has a total blood volume of 5.0L. What volume (in mL) of 6.0 M HC

Question

A 70-kg person has a total blood volume of 5.0L. What volume (in mL) of 6.0 M HCl could be neutralized by blood without the blood pH dropping below 7.0? Carbonic acid has a pKa of 6.1, and a pH of 7.4. The concentration of Carbonic acid in human blood is .0012M. The concentration of its conjugate base (bicarbonate ion) in human blood is .024M.

I worked it out like this, but am unsure if it's right.....

Acid: [0.0012](5.0L)=0.006 mol carbonic acid
Base: [0.024](5.0L)=0.12 mol bicarbonate base <--limiting

So, Volume HCl=(0.12 mol bicarbonate/6.0 M HCl)=0.02L HCl=20mL HCl

Is this right? We didn't really take this approach to solving anything in class, but the units seem to cancel out to leave me with the amount of HCl that the bicarbonate could neutralize.

Explanation / Answer

pH = pKa + log(A/HA) Let's keep everything in moles for now, then find volume If pKa of carbonic acid is 6.1 and the minimum allowed pH is 7, and concentration of HA is 0.0012M then... pH = pKa + log [HCO3-]/ [H2CO3] 7.0 = 6.1 + log [HCO3-]/ [H2CO3] 7.0 - 6.1 = 0.9 HCO3- + H+ = H2CO3 10^0.9=7.94 = [HCO3-]/ [H2CO3]= 0.024-x/ 0.0012+x 0.00953 + 7.94x= 0.024- x 8.94 x =0.0145 x = 0.00162 M = concentration HCl needed Thats how much HCl the blood can tolerate, so now we find volume volume of HCl needed = moles needed / concentration = X / 6 moles needed = 0.00162M HCl in blood * 5L of blood = 0.0081 mol Volume needed = 0.0081 mol / 6M = 0.00135L = 1.35 mL HCl Can't use your approximation because pKa of the weak acid (carbonic acid) must be taken into account. Since it is a weak acid then a significantly smaller amount of strong acid (HCl) is needed.

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