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Bess [88]
3 years ago
15

Which buffer system is found in the human body?

Chemistry
1 answer:
aivan3 [116]3 years ago
8 0

The correct answer is the bicarbonate buffer system.  

The bicarbonate buffer system refers to an acid-base homeostatic mechanism taking part in the balance of bicarbonate ion, carbonic acid, and carbon dioxide in order to sustain the pH in the duodenum and blood, among other tissues, to support the entire metabolic activities.  

The bicarbonate buffer system plays a vital role in the human body systems. In the duodenum and the stomach, the bicarbonate buffer system functions to both neutralize gastric acid and steady the intracellular pH of epithelial cells through the secretion of bicarbonate ion into the gastric mucosa.  


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The balanced reaction is:

N2 + 3H2 = 2NH3

We are given the amount of the product to be produced.This will be the starting point of our calculations. We use the ideal gas equation to find for the number of moles.

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n = PV / RT = 1.00(.520 L) / (0.08206 atm L/mol K ) 273 K
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An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni
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<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}    .....(1)

  • <u>For KOH:</u>

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol

  • <u>For propanoic acid:</u>

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol

The chemical reaction for propanoic acid and KOH follows the equation:

                 C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           0.09206          -                0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L     (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})

pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})

We are given:  

pK_a = negative logarithm of acid dissociation constant of propanoic acid = 4.89

[C_2H_5COOK]=\frac{0.04514}{0.26594}

[C_2H_5COOH]=\frac{0.09206}{0.26594}

pH = ?  

Putting values in above equation, we get:

pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58

Hence, the pH of acid solution is 4.58

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