Blood is buffered by carbonic acid and the bicarbonate ion. Normal blood plasma is 0.024 Min HCO?3 and 0.0012 M H2CO3 (pKa1 for H2CO3 at body temperature is 6.1).

A. What is the pH of blood plasma?

I got 7.4 for pH which is the correct answer
B. If the volume of blood in a normal adult is 5.0 L, what mass of HCl could be neutralized by the buffering system in blood before the pH fell below 7.0 (which would result in death)?
C. Given the volume from part B, what mass of NaOHcould be neutralized before the pH rose above 7.8?

Express your answer using two significant

Blood is buffered by carbonic acid and the bicarbonate ion. Normal blood plasma is 0.024 M in HCO?3 and 0.0012 M H2CO3. The pH of the blood plasma is 7.4, the mass of HCl = 0.3 grams, and the mass of NaOH = 0.14 g

From the given information;

The dissociation of carbonic acid in water results in the formation of bicarbonate ion and hydronium ion.

∴

By applying Henderson equation;

pH of blood plasma = 7.4

B.

From the given question, let (x) be the amount of HCl that was added, then the salt reacts back to produce carbonic acid again.

By the application of the Henderson-Hasselbalch equation:

Using logarithm rules;

(0.12 - x) = 7.94(0.006 + x)

0.12 - x = 0.04764 + 7.94x

0.12 - 0.04764 = 7.94x +x

0.07236 = 8.94x

x = 0.07236/8.94

x = 0.008 moles

Recall that:

number of moles = mass/molar mass

C.

Using the 5.0 L of blood that a normal human has in part B;

The number of moles of  the buffer component can be computed as:

Suppose, (x) moles of NaOH could be neutralized by H2CO3;

Then, at equilibrium:

By the application of the Henderson-Hasselbalch equation:

Making (x) the subject of the formula:

Recall that:

number of moles = mass/molar mass

For NaOH,

mass = number of moles of NaOH × molar mass

mass of NaOH = (0.0035 moles × 40 g/mol )

mass of NaOH = 0.14 g

Therefore, we can conclude that the pH of the blood plasma is 7.4, the mass of HCl = 0.3 grams, and the mass of NaOH = 0.14 g

Learn more about the Henderson-Hasselbalch equation here:

link

See explanations

Explanation:

a. pH = Pka + log([Conj Base]/[Wk Acid])

pH = 6.1 + log(0.024/0.0012) = 6.1 + 1.3 = 7.4

b. Excess acid is removed from system by combining with conjugate base causing the buffer reaction to shift left forming more H₂CO₃.

Given 0.024M HCO₃⁻ in buffer, 0.024 mole HCl would be removed => mass HCl removed = (0.024mol)(36 g/mol) = 0.864 g HCl.

c. Excess OH⁻ (strong base) is removed when OH⁻ combines with H⁺ forming water. This causes more H₂CO₃ to decompose (rxn shifts right) and continue until all of the OH⁻ is removed.

Given 0.0012M H₂CO₃ in buffer, then the max amount of NaOH that can be removed is (0.0012mol)(40 g/mol) = 0.048 gram NaOH removed.  

Another way to visualize the action of the buffer system is to place the reaction on a see-saw and note which way if tilts when a stress is applied to the equilibrium. The chemistry will then take over and shift the reaction to establish a new equilibrium with new concentration values to the limit of the  available reagents in the system.

                                           H₂CO₃ <=> H⁺ + HCO₃⁻

Adding excess H⁺ adds weight on product side & tilts reaction see-saw right => then reaction shifts left to establish new equilibrium.  

Adding excess OH⁻ causes H⁺ to combine with OH⁻ forming water essentially removing OH⁻ from the equilibrium see-saw (lighter on product side) => rxn see-saw tilts left then shifts right to establish the new equilibrium.

Both processes proceed to the limit of the reagent concentrations making up the buffer solution.

For a buffer system with 0.0012 H₂CO₃ & 0.024M HCO₃⁻

=> adding acid will cause 0.024 mole of the HCO₃⁻ to combine with 0.024 mole of H⁺ (HCl in this case) => 0.024 mole x 36 g/mole = 0.864 g. HCl max.

=> adding base will cause 0.0012 mole H₂CO₃ to decompose (right shift) to replace 0.0012 mole OH⁻ that was converted to H₂O on combining with H⁺ of buffer => 0.0012 mol x 40 g/mol = 0.048 gram NaOH neutralized.

- decrease the number of plants and animals living in the water.

2b2t

explanation:

2b2t