Question #42665

In 10 mL of blood sample pH was 7,56. A high amount of acid was added into the blood. After that volume of CO2 in a sample was 6,13mL. pKa is 6,1. Find concentrations of CO2 and HCO3- in blood.

Please, help me, I almost understand how everything should be calculated, but somewhere I keep thinking faulty I guess because can't get the possible answers anyhow...

Please, help me, I almost understand how everything should be calculated, but somewhere I keep thinking faulty I guess because can't get the possible answers anyhow...

Expert's answer

Find the amount of substance of CO_{2}. Assume that the volume was measured at ambient conditions (298.15 K, 100 kPa).

n(CO_{2}) = pV/RT = 2.473*10^{-4} mol

Assume that the only substance responsible for blood pH is CO_{2}. Carbon dioxide dissociates in water according to the following equation:

CO_{2} + H_{2}O --> H^{+} + HCO_{3}^{-}

We have a buffer system, and its pH can be calculated from the following equation:

pH = pK_{a} + lg [A^{-}]/[HA]

where [A^{-}] is the concentration of bicarbonate ion HCO_{3}^{-}, and [HA] is the concentration of free CO_{2}.

[A^{-}]/[HA] = 10^(pH - pK_{a}) = 28.84

The total amount of CO_{2} and HCO_{3}^{-} in the solution is equal to the amount of evolved CO_{2}. Therefore we can find the total concentration of these particles:

c(CO_{2} + HCO_{3}^{-}) = n(CO_{2})/V(solution) = 2.473*10^{-4}/0.01 = 0.02473 mol/L

Let x denote the concentration of free CO_{2} in the solution. Hence the concentration of HCO_{3}^{-} ions is 28.84x

x + 28.84x = 0.02473

x = 8.29*10^{-4} mol/L

28.84x = 2.39*10^{-2} mol/L

Answer:

c(CO_{2}) = 8.29*10^{-4} mol/L

c(HCO_{3}^{-}_{}) = 2.39*10^{-2} mol/L

n(CO

Assume that the only substance responsible for blood pH is CO

CO

We have a buffer system, and its pH can be calculated from the following equation:

pH = pK

where [A

[A

The total amount of CO

c(CO

Let x denote the concentration of free CO

x + 28.84x = 0.02473

x = 8.29*10

28.84x = 2.39*10

Answer:

c(CO

c(HCO

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