Question

A human patient suffering from an ulcer may show a hydrochloric acid concentration of 0.075mol/L in his gastric juice. It is possible to neutralise this acid with aluminium hydroxide $Al{(OH)_3}$ which reacts with HCl according to the chemical reaction shown below: [Al = 27]
                                          $Al{(OH)_3} + HCl \to AlC{l_3} + {H_2}O$
Find the weight of $Al{(OH)_3}$ which is required to neutralise 2 litre gastric acid.
(A) 1.95g
(B) 6.20g
(C) 0.693g
(D) 3.9g

Answer 1

Hint: First find out the amount of HCl present in 2L gastric acid. Then write down the balanced chemical reaction for this neutralisation. Now use this reaction and the unitary method to find out the amount of $Al{(OH)_3}$ which is required to neutralise the amount of HCl present in 2 litre gastric acid.

Complete step by step answer:
-According to the question the amount of HCl present in the stomach is 0.075 mole in 1L gastric juice and we need to find out the weight of $Al{(OH)_3}$ which is required to neutralise 2 litre HCl (gastric juice).
-So, we will first find out the weight of HCl present in 2L of gastric juice.
If 1L of gastric juice contains 0.075 moles of HCl then the moles of HCl present in 2L of gastric juice = 0.075 × 2 = 0.15
Molar mass of HCl is = 36.5 g/mol
And the weight of 0.15 moles of HCl will be = 0.15 × 36.5 = 5.475 g

-The balanced chemical equation involved in the neutralisation of HCl by $Al{(OH)_3}$is:
$Al{(OH)_3} + 3HCl \to AlC{l_3} + 3{H_2}O$
According to this reaction: 1 mole of $Al{(OH)_3}$is neutralised by 3 moles of HCl.
Molecular weight of $Al{(OH)_3}$= 78 g/mol
Weight of 3 moles of HCl = 3 × 36.5 = 109.5 g
So, we can also say that: 109.5 g of HCl will neutralise 78 g of $Al{(OH)_3}$.

We will now use the unitary method to find out the amount of $Al{(OH)_3}$ which will be neutralised by 109.5 g HCl neutralise → 78 g $Al{(OH)_3}$
1 g HCl will neutralise → $\dfrac{{78}}{{109.5}}$ g $Al{(OH)_3}$
5.475 g HCl will neutralise → $\dfrac{{78}}{{109.5}} \times 5.475$

= 3.9 g of $Al{(OH)_3}$

So, the amount of $Al{(OH)_3}$which is required to neutralise 2L of gastric acid will be 3.9 g.

So, the correct answer is “Option D”.

Note: Gastric juice is a digestive fluid which is formed inside the lining of the stomach. The parietal cells present in the gastric glands in the stomach secrete this HCl. The HCl present in the gastric juice activates the digestive enzymes present in it which promote digestion of proteins. They break the long chains of amino acids which form the protein molecules. The pH of the stomach due to HCl is regulated by bicarbonate ($NaHC{O_3}$) secreted by the stomach.