# Finding concentration of hydronium ions in water after adding hydrogen chloride gas [closed]

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.

• Did you forget to complete the attempt, or is this it? Sep 19 '20 at 5:58
• @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. Sep 19 '20 at 6:04
• 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? Sep 19 '20 at 6:07
• @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. Sep 19 '20 at 6:20
• 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. Sep 19 '20 at 6:21

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).

• 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? Sep 19 '20 at 8:29
• That is the key ($\ce{H+ + H2O -> H3O+}$). Sep 19 '20 at 8:32
• Sorry, I am beginner in chemistry study. So i don't know what is $\ce{HH3}$ and $\ce{HH4+}$? Would you tell me? Sep 19 '20 at 8:41
• See equation (3) under notes (similar to $\ce{H2O}$ and $\ce{H3O+}$). Sep 19 '20 at 8:52
• 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? Sep 19 '20 at 12:45