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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

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ajeigbeibraheem

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;

  • pKa1 for H2CO3(carbonic acid) at body temperature is 6.1
  • Normal blood plasma in HCO₃⁻ = 0.024 M
  • Normal blood plasma in H2CO3 = 0.0012 M

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

∴

[tex]\mathbf{H_2CO_3_{(aq)} + H_2O \to H_2CO_3^-_{(aq)}+ H_3O^+_{(aq)}}[/tex]

By applying Henderson equation;

[tex]\mathbf{pH = pKa_1 + log \dfrac{[HCO_3^-]}{[H_2CO_3]}}[/tex]

[tex]\mathbf{pH =6.1 + log \dfrac{[0.024]}{[0.0012]}}[/tex]

[tex]\mathbf{pH =6.1 + 1.30}}[/tex]

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:

[tex]\mathbf{pH = pKa_1+ log \dfrac{[ salt]}{[acid] }}[/tex]

[tex]\mathbf{7 =6.1+ log \dfrac{[ 0.024 \times 5 -x]}{[0.0012 \times 5 +x] }}[/tex]

[tex]\mathbf{0.9= log \dfrac{[ 0.024 \times 5 -x]}{[0.0012\times 5 +x] }}[/tex]

Using logarithm rules;

[tex]\mathbf{10^{0.9}= \dfrac{[ 0.024 \times 5 -x]}{[0.0012 \times 5 +x] }}[/tex]

[tex]\mathbf{7.94 = \dfrac{[ 0.024 \times 5 -x]}{[0.0012 \times 5 +x] }}[/tex]

[tex]\mathbf{7.94 = \dfrac{[ 0.12-x]}{[0.006 +x] }}[/tex]

(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

  • mass of HCl = number of moles of HCl × molar mass
  • mass of HCl = 0.008 moles × 36.5 g/mol
  • mass of HCl = 0.3 grams

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:

[tex]\mathbf{n_{HClO_3^-} = 5 \ L ( 0.024 \ mol/L)}} \\ \\ \\ \mathbf{ n_{HClO_3^-} = 0.12 mol}[/tex]

[tex]\mathbf{n_{H_2CO_3} = 5 \ L ( 0.0012 \ mol/L)}} \\ \\ \\ \mathbf{ n_{H_2CO_3}= 0.0060 \ mol}[/tex]

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

Then, at equilibrium:

  • [tex]\mathbf{n_{HClO_3^-} =0.12 + x}}[/tex]
  • [tex]\mathbf{n_{H_2CO_3} =0.0060 - x}}[/tex]

By the application of the Henderson-Hasselbalch equation:

[tex]\mathbf{pH = pKa_1+ log \dfrac{[ HCO_3^-]}{[H_2CO_3] }}[/tex]

[tex]\mathbf{7.4 = 6.1+ log \dfrac{0.12 +x}{0.0060-x} }[/tex]

Making (x) the subject of the formula:

[tex]\mathbf{1.3= log \dfrac{0.12 +x}{0.0060-x} }[/tex]

[tex]\mathbf{x = 0.0035 \ moles}[/tex]

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:

https://brainly.com/question/13423434?referrer=searchResults

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