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A sample of hydrogen chloride gas, $\ce{HCl}$, occupies $\pu{0.932 l}$ at a pressure $\pu{1.44 bar}$ and at a temperature of $\pu{50^\circ C}$. The sample is dissolved in $\pu{1 l}$ of water.
What is the resulting hydronium ion ($\ce{H3O+}$) concentration?

Ammonia gas also dissolved quantitatively in water.
If it is measured at $\pu{0.720 bar}$ and $\pu{50^\circ C}$, what volume of $\ce{NH3}$ gas is required to neutralize the solution prepared in the above question?

For present purposes, assume that the neutralization reaction occurs quantitatively.

My attempt:

$\pu{0.05 mol}$ of $\ce{HCl}$ is dissolved in the $\pu{55.55 mol}$ of water. Balancing equation is

$$\ce{HCl(g) + H2O(l) <=> H3O+ + Cl-(aq)}$$

Is there a mistake in this assumption? I ask this because I am unable to proceed further.

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  • $\begingroup$ Did you forget to complete the attempt, or is this it? $\endgroup$ Sep 19 '20 at 5:58
  • $\begingroup$ @Safdar, I didn't forget to complete the attempt. I just don't have any idea to proceed further. I know molarity formula, but i am doubtful about its usefulness here. $\endgroup$ Sep 19 '20 at 6:04
  • $\begingroup$ I've removed the second part of your question since it is basically the same thing but using $\ce{OH-}$ instead.. Also, HCl is a strong electrolyte. Does this help? $\endgroup$ Sep 19 '20 at 6:07
  • $\begingroup$ @Safdar, I think the second part is required because the author is asking what volume of ammonia gas $\ce{NH3}$ is necessary to neutralize the solution created in first part. $\endgroup$ Sep 19 '20 at 6:20
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    $\begingroup$ Is there a reason to suppose you get negatively charged gas consisting of chloride ions, while solution would be charged positively ? BTW, I replaced the character you have used by the regular minus to be displayed properly. $\endgroup$
    – Poutnik
    Sep 19 '20 at 6:21
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You correctly found that amount of $\ce{HCl}$ gas in the container is $\pu{0.05 mol}$. When it dissolves in $\pu{1.0 L}$ of water you get $\pu{0.05 M}$ $\ce{HCl}$ solution. Does it matter for this calculations? No, but I did it anyway. However, it gives us very important information. $\ce{HCl}$ is a strong electrolyte and it dissolves in water as high as $\pu{12 M}$. When dissolve in water, it completely dissociates:

$$\ce{HCl (g) + H2O (l) -> H3O+ (aq) + Cl- (aq)} \tag1$$

For the second part of the question, you need to know what is the reaction between $\ce{HCl}$ and $\ce{HH3}$:

$$\ce{HCl (aq) + NH3 (aq) -> NH4Cl (aq) -> NH4+ (aq) + Cl- (aq)} \tag2$$

Thus, mol% of the reaction is $1:1$. Therefore, the amount of $\ce{HH3 (g)}$ you need to complete neutralization is the amount of $\ce{HCl}$ in the solution, which is $\pu{0.05 mol}$. Now, you can use ideal gal law to calculate the $V$ under given conditions.


Note: Ammonia in water is in equilibrium with: $$\ce{NH3 (g) + H2O (l) <=> NH4+ (aq) + OH- (aq)} \tag3$$ The complete nutralization is the reactions of $(1)$ and $(3)$ where net ionic reaction is: $$\ce{H3O+ (aq) + OH- (aq) -> 2 H2O (l)} \tag4$$

However, since $\ce{NH3}$ is a weak base, its conjgate acid $\ce{NH4+}$ is acidic. Therefore, this titration would not pass $\mathrm{pH = 7}$ mark at the end point (thus need an appropriate indicator).

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  • $\begingroup$ So, the concentration of $\ce{H3O+}$ in the solution of HCL created in first part is $\pu{0.05 mol l^{-1}}$. Isn't it? $\endgroup$ Sep 19 '20 at 8:29
  • $\begingroup$ That is the key ($\ce{H+ + H2O -> H3O+}$). $\endgroup$ Sep 19 '20 at 8:32
  • $\begingroup$ Sorry, I am beginner in chemistry study. So i don't know what is $\ce{HH3}$ and $\ce{HH4+}$? Would you tell me? $\endgroup$ Sep 19 '20 at 8:41
  • $\begingroup$ See equation (3) under notes (similar to $\ce{H2O}$ and $\ce{H3O+}$). $\endgroup$ Sep 19 '20 at 8:52
  • $\begingroup$ So, the volume of ammonia gas $\ce{NH3}$ required to neutralize the solution prepared in the first part at 0.720 bar and $50^\circ C$ is 1.866 liters. Isn't it? $\endgroup$ Sep 19 '20 at 12:45

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