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AnAn

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Q1. arrange the following gas mixtures in order of their tendency for CO2(g) to dissolve in aqueous solution ( percentages given are mole %)
1) 90% Ar, 10% CO2
2) 80% Ar, 10% CO2, 10% NH3
3) 80% Ar, 10% CO2, 10% Cl2
Write equaions for any chemical reactions occurring in aqueous solution on exposure to these gas mixtures.

Q2. the demonstration involves placing 100mL of distilled water, 100mL of 1.0M acetic acid and 100 mL of 1.0M hydrochloric acid respectively in three 250mL beakers. An electical circuit with a power source and light bulb is set up, the solution in the beakers acting to complete the circuit.

a) to the surprise of students the light bulb over the beaker with water glowed faintly indicating the water was conducting electricity, Explain this phenomenon.

b) predict the difference in brightness of the three light bulbs
 

Xayma

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Q)1)

2) Is top, as NH<sub>3</sub> will form a weak base in the water which will react with the carbonic acid formed from the CO<sub>2</sub> hence more will dissolve.

I think it would be 1) then 3) next dont ask me why its just a gut feeling.
 

speed_bump

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Q2. a) because the water isnt pure and would contain dissoved ions...

b)water- faintly
acetic acid- stronger
hydochloric- strongest

i think.... acetic acid is weaker then hydrocloric right??
 

CM_Tutor

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Xayma, I think you're right, as I suspect that chlorine will undergo a similar hydrolysis to that of bromine, and thus will generate a slightly acidic solution.

Speed_bump, even pure water contains ions due to the auto-ionisation of water:

2H<sub>2</sub>O<sub> (l)</sub> <---> H<sub>3</sub>O<sup>+</sup><sub>(aq)</sub> + OH<sup>-</sup><sub>(aq)</sub>

and this is the focus of the question, IMO.

And yes, acetic acid weak whereas hydrochloric acid is strong, and thus the brightest light will be related to the hydrochloric acid solution.
 

AnAn

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thx for help~~
i got one more question to ask.....

1. Write equations for the reactions taking place when carbon dioxide is dissolve in aqueous sodium hydroxide solution.
 

CM_Tutor

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Start with the usual equations for dissolution of carbon dioxide in water:

CO<sub>2 (g)</sub> <---> CO<sub>2 (aq)</sub>

CO<sub>2 (aq)</sub> + H<sub>2</sub>O<sub> (l)</sub> <---> H<sub>2</sub>CO<sub>3 (aq)</sub>

H<sub>2</sub>CO<sub>3 (aq)</sub> + H<sub>2</sub>O<sub> (l)</sub> <---> H<sub>3</sub>O<sup>+</sup><sub>(aq)</sub> + HCO<sub>3</sub><sup>-</sup><sub>(aq)</sub>

HCO<sub>3</sub><sup>-</sup><sub>(aq)</sub> + H<sub>2</sub>O<sub> (l)</sub> <---> H<sub>3</sub>O<sup>+</sup><sub>(aq)</sub> + CO<sub>3</sub><sup>2-</sup><sub>(aq)</sub>

In addition, the carbonic acid can react directly with the hydroxide present

OH<sup>-</sup><sub>(aq)</sub> + H<sub>2</sub>CO<sub>3 (aq)</sub> <---> HCO<sub>3</sub><sup>-</sup><sub>(aq)</sub> + H<sub>2</sub>O<sub>(l)</sub>

as can the hydrogencarbonate ions

OH<sup>-</sup><sub>(aq)</sub> + HCO<sub>3</sub><sup>-</sup><sub>(aq)</sub> <---> CO<sub>3</sub><sup>2-</sup><sub>(aq)</sub> + H<sub>2</sub>O<sub>(l)</sub>

And also, the hydroxide will react with the hydronium produced by the carbonic acid:

OH<sup>-</sup><sub>(aq)</sub> + H<sub>3</sub>O<sup>+</sup><sub>(aq)</sub> <---> 2H<sub>2</sub>O<sub>(l)</sub>

More CO<sub>2</sub> will dissolve in aqueous sodium hydroxide because these last three reactions cause the second, third and fourth reactions to move right (by applying Le Chatelier's Principle), and thus the first reaction also moves right (Le Chatelier's Principle), thereby increasing the solubility of carbon dioxide.
 
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