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meriva
3 years ago
9

The main buffering agents in human blood are h2co3 and hco3. Explain how this buffering system will be able to maintain blood pH

at a steady level when a small amount of a generic acid HA ( with ions H+ and A-) is added
Chemistry
1 answer:
den301095 [7]3 years ago
6 0

Answer:

The base, HCO₃⁻ of the buffering system will neutralize the acid added forming H₂CO₃.

Explanation:

A buffer solution keeps the pH in a narrow range upon the addition of an acid or base. A buffer consists of a pair of a weak acid and its conjugate base, or vice versa, in a pair of a weak base and its conjugate acid. In the case of human blood, the acid is H₂CO₃ and its conjugate base is HCO₃⁻.

The buffering system of the human blood maintain blood pH at a steady level when a small amount of a generic acid HA is added in the following way:

HCO₃⁻ + HA  ⇄  H₂CO₃ + A⁻

The base, HCO₃⁻, will neutralize the acid added forming the H₂CO₃, which is already a component of the buffering system of the human blood. Thus, hydronium ions of the acid added are removed, preventing the pH of blood from becoming acidic.                

On the other hand, when a small amount of a generic base A⁻ is added the following reaction takes place:    

H₂CO₃ + A⁻  ⇄  HCO₃⁻ + HA

In this case, the acid H₂CO₃ will neutralize the base added to form the base HCO₃⁻, which is already a component of the buffering system. Thus, the base added are removed from the blood, preventing the pH of blood from becoming basic.  

Since, the acid or its conjugate base will react with the base or acid added, to neutralizing it and forming the species that conform the buffer solution, the pH is being maintained at a steady level.  

In the process above, the addition of a small amount of an acid or a base won't change in a significant way the concentrations of the two components of the buffering system. This buffering mechanism prevents the blood from becoming acidic or basic.  

                   

I hope it helps you!

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Answer:

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Explanation:

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3 years ago
Determine the molarity for each of the following solution solutions:
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Answer :

(a)The molarity of KCl solution is, 0.9713 mole/L

(b)The molarity of H_2SO_4 solution is, 0.00525 mole/L

(c)The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

(d)The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

(e)The molarity of Br_2 solution is, 0.0565 mole/L

(f)The molarity of C_2H_5NO_2 solution is, 0.0113 mole/L

Explanation :

<u>(a) 1.457 mol of KCl in 1.500 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Solute is KCl.

\text{Molarity of the solution}=\frac{1.457mole}{1.500L}=0.9713mole/L

The molarity of KCl solution is, 0.9713 mole/L

<u>(b) 0.515 gram of H_2SO_4, in 1.00 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is H_2SO_4

Molar mass of H_2SO_4 = 98 g/mole

\text{Molarity of the solution}=\frac{0.515g}{98g/mole\times 1.00L}=0.00525mole/L

The molarity of H_2SO_4 solution is, 0.00525 mole/L

<u>(c) 20.54 g of Al(NO_3)_3 in 1575 mL of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Solute is Al(NO_3)_3

Molar mass of Al(NO_3)_3 = 213 g/mole

\text{Molarity of the solution}=\frac{20.54g\times 1000}{213g/mole\times 1575L}=0.0612mole/L

The molarity of Al(NO_3)_3 solution is, 0.0612 mole/L

<u>(d) 2.76 kg of CuSO_4.5H_2O in 1.45 L of solution</u>

Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Solute is CuSO_4.5H_2O

Molar mass of CuSO_4.5H_2O = 250 g/mole

\text{Molarity of the solution}=\frac{2760g}{250g/mole\times 1.45L}=7.61mole/L

The molarity of CuSO_4.5H_2O solution is, 7.61 mole/L

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Formula used :

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}

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\text{Molarity of the solution}=\frac{0.005653mole\times 1000}{10.00L}=0.0565mole/L

The molarity of Br_2 solution is, 0.0565 mole/L

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Formula used :

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

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Molar mass of C_2H_5NO_2 = 75 g/mole

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