Acetylsalicylic acid (C_9H_8O_4) commonly known as aspirin. has been used for ov
ID: 528832 • Letter: A
Question
Acetylsalicylic acid (C_9H_8O_4) commonly known as aspirin. has been used for over a hundred years as an effective pain reliever. When aspirin reaches the blood stream. it dissociates into H+ and its conjugate base, effectively decreasing the pH of blood: HA rightleftarrow H^+ + A^- A person may go into shock or die if blood pH drops below 6.8. However buffers in the blood help dampen changes in blood pH. The main buffer is the bicarbonate system: H_2CO_3 rightleftarrow H^+ + HCO^-_3 When aspirin dissociates, the bicarbonate equilibrium shifts towards carbonic acid (H_2CO_3). thus reducing the concentration of H+ in the blood stream. The initial concentration of H_2CO_3 and HCO_3, in the blood stream is 0.0014 mol/L and 0.0266 mol/L. respectively. The aspirin equilibrium has the bicarbonate equilibrium has pk_a = 636. Assume that the blood volume in the body is 5 Land the molecular weight of the acetylsalicylic acid is 1802 g/mol. Calculate the equilibrium constants (K_a) for the two dissociation reactions. Write molar balances for the species involved in the bicarbonate equilibrium. Calculate the moles of carbonic acid at equilibrium when the patient goes into shock (pH 6.8). Write molar balances for the species involved in the acetylsalicylic acid equilibrium. Calculate the mass of aspirin to be consumed for the patient to go into shock(pH = 6.8).Explanation / Answer
3a.
Ka1 = 10^-pKa1= 10^-3.6 = 0.0002511
Ka2 = 10^-pKa2= 10^-6.36 = 4.365*10^-7
3b
molar balance:
bicarbonate -- HCO3-
HCO3-(aq) = H+(aq) + CO3-2(aq)
H2CO3(aq) <--> H+(aq) + HCO3-(aq)
3c.
molse of H2CO3 in equilbirium withj ´H = 6.8
pH = pKa+ log(HCO3-/H2CO)
initially
H2CO3 = 0.0014
HCO3- = 0.0266
6.8 = 6.6 + log(0.0266 + x / 0.0266 -x)
10^(6.8-6.6) = (0.0014+ x / 0.0266 -x)
1.584*( 0.0266 -x) = (0.0014+ x
1.584*( 0.0266) - x*1.584 = 0.0014 +x
1.584*( 0.0266)-0.0014 = (1+1.584)x
x = (1.584*( 0.0266)-0.0014 )/( (1+1.584))
x= 0.01576
[H2CO3] = 0.0266 -0.01576 = 0.01084 M
V = 5 L
mol = MV = 5*0.01084= 0.0542 mol of H2CO3
d)
molar balance for species in aspirin:
C8H7O-COOH <-> H+ + C8H7O-COO-