Question: The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO2), is partially regulated by the concentration of H3O+ and dissolved CO2 in the blood. Although the equilibrium is complicated, it can be summarized as

HbO2(aq) + H3 O+(aq) + CO2(g) ⇌ CO2 −Hb−H+ + O2(g) + H2 O(l)

(a) Write the equilibrium constant expression for this reaction.

(b) Explain why the production of lactic acid and CO2 in a muscle during exertion stimulates release of O2 from the oxyhemoglobin in the blood passing through the muscle.

Short Answer

Expert verified

The Result is

  1. \({K_c} = \frac{{\left( {{\rm{C}}{{\rm{O}}_2} - {\rm{Hb}} - {{\rm{H}}^ + }} \right) \times \left( {{{\rm{O}}_2}} \right)}}{{\left( {{\rm{Hb}}{{\rm{O}}_2}} \right) \times \left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right) \times \left( {{\rm{C}}{{\rm{O}}_2}} \right)}}\)The Expression of the equilibrium constants is
  2. The Increase in concentration of lactic acid \(\left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right)\)and CO2will move equilibrium in the right, therefore oxyhemoglobin HbO2releases O2

Step by step solution

01

Definition

Oxyhaemoglobin is the haemoglobin bound to oxygen and oxygen is transported in this form to tissues from the lungs

02

Expression for the equilibrium constant

The Equillibrium constant expression is

\({K_c} = \frac{{\left( {{\rm{C}}{{\rm{O}}_2} - {\rm{Hb}} - {{\rm{H}}^ + }} \right) \times \left( {{{\rm{O}}_2}} \right)}}{{\left( {{\rm{Hb}}{{\rm{O}}_2}} \right) \times \left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right) \times \left( {{\rm{C}}{{\rm{O}}_2}} \right)}}\)

03

Step 3:  The Release of O2 from the oxyhemoglobin

We Have to explain the production of lactic acid and CO2in the muscle during the exertion stimulates the release the O2 from the oxyhemoglobin in the blood passing through the muscle.

The Increase in concentration of lactic acid\(\left( {{{\rm{H}}_3}{{\rm{O}}^ + }} \right)\)andCO2will move equilibrium in the right, therefore oxyhemoglobin HbO2releases O2

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Most popular questions from this chapter

Question:Calculate the value of the equilibrium constant \({K_P}\) for the reaction \(2NO(g) + C{l_2}(g) \rightleftharpoons 2NOCl(g)\) from these equilibrium pressures: NO, \(0.050atm;C{l_2},0.30atm;NOCl,1.2atm\)

For which of the reactions in Exercise 13.15 does\({K_c}\)(calculated using concentrations) equal\({K_p}\)(calculated using pressures)?

(a) \(C{H_4}(g) + C{l_2} \rightleftharpoons C{H_3}CI(g) + HCI(g)\)

(b) \({N_2}(g) + {O_2}(g)\rightleftharpoons 2NO(g)\)

(c) \(2S{O_2}(\;g) + {O_2}(\;g)\rightleftharpoons 2S{O_3}(\;g)\)

(d) \(BaS{O_3}(s)\rightleftharpoons BaO(s) + S{O_2}(g)\)

(e) \({P_4}(g) + 5{O_2}(g)\rightleftharpoons{P_4}{O_{10}}(s)\)

(f) \(B{r_2}(\;g)\rightleftharpoons 2Br(g)\)

(g) \(C{H_4}(g) + 2{O_2}(g)\rightleftharpoons C{O_2}(g) + 2{H_2}O(l)\)

(h)\(CuS{O_4} \times 5{H_2}O(s)\rightleftharpoons CuS{O_4}(s) + 5{H_2}O(g)\)

If you observe the following reaction at equilibrium, is it possible to tell whether the reaction stated with pure \(N{O_2}\) or with pure \({N_2}{O_4}\)? \(2N{O_2}(g) \rightleftharpoons {N_2}{O_4}(g)\)

Question: The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose.

C12 H22 O11(aq) + H2O(l)⟶C6 H12 O6 (aq) + C6 H12 O6 (aq)

Rate = k[C12H22O11]

In neutral solution, k = 2.1 × 10−11/s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C6H12O6, but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 105 at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 M aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1.

Complete the changes in concentrations (or pressure, if requested) for each of the following reactions.

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