Is this problem right

Blood has a normal pH of 7.35-7.45 and contains two major buffer systems. It is important that the pH of blood remains relatively constant because at pH below 6.8 or greater than 8.0, cells cannot function properly and death may result. The HCO3-/CO2 (aq) blood buffer in vivo is an open system in which the concentration of dissolved CO2 is maintained constant. Any excess CO2 produced by the reaction H+ + HCO3- ---> H2O + CO2 is expelled by the lungs. Note that a typical laboratory buffer is a closed system. The concentration of conjugate acid increases when H+ reacts with the conjugate base.
You calculated the Keq and pK of Reaction (4) from the following reactions and K values in lecture.
CO2 (g) <==> CO2 (aq) K1 = 3 x 10-5 at 37oC.
CO2 (aq) + H2O (l) <==> H2CO3 (aq) K2 = 5 x 10-3 at 37oC
H2CO3 (aq) <==> H+ (aq) + HCO3- (aq) pKa = 3.8 at 37oC
CO2 (aq) + H2O (l) <==> H+ (aq) + HCO3- (aq) K4 = ?
You calculated the [HCO3-] = 0.024 M in blood at pH 7.4. Calculate the [CO2 (aq)] in blood at this pH.
0.01 M H+ is added to blood. You calculated the pH of blood under conditions such that the increased [CO2 (aq)] can be released as CO2 (g). In other words, assume that the blood buffer is an open system. Remember that the [CO2 (aq)] remains constant in this open buffer system.
a. Cells cannot function property if the pH of blood falls below 6.8 or rises above 8.0. Calculate the amount in M of H+ that is added to blood for the blood pH to fall to 6.8.

Here's my work,
CO2 (aq) + H2O → H+ + HCO-3
initial .0012 .018 M .024
reacted - 0 added -.018 M
equillibrium .0012 M .006 M

PH= 6.1 + log (.006 M/ .0012 M)= 6.8 PH

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