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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: T on April 08, 2016, 02:50:04 AM
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An aqueous solution of ammonia has of pH = x, and a solution of hydrochloric acid has pH = y; it is also known that x + y = 14, and x > 11. If equal volumes of these two solutions are mixed together, what would be the concentration of the various ions in the resulting solution in descending order?
A [NH4+] > [Cl–] > [OH–] > [H+]
B [Cl–] > [NH4+] > [H+] > [OH–]
C [NH4+] > [Cl–] > [H+] > [OH–]
D [Cl–] > [NH4+] > [OH–] > [H+]
E [Cl–] = [NH4+] = [OH–] = [H+]
Since x+y = 14 then there are equal amounts of OH- and H+ to start with. To prove this:
x = 14 - y
[H+] of ammonia solution is 10-y
[OH-] of ammonia solution is 10-14/10-y = 10y-14
[H+] of hydrochloric acid solution is 10-x=10y-14
Hence, [OH-] in ammonia = [H+] in hydrochloric acid solution then that means there is a lot more ammonia then hydrochloric acid since ammonia is a weak base. Therefore there the answer is either A or C. However, I cannot decide between those two, can someone give me hints?
Thanks
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An aqueous solution of ammonia has of pH = x, and a solution of hydrochloric acid has pH = y; it is also known that x + y = 14, and x > 11. If equal volumes of these two solutions are mixed together, what would be the concentration of the various ions in the resulting solution in descending order?
Since x+y = 14 then there are equal amounts of OH- and H+ to start with. To prove this:
x = 14 - y
[H+] of ammonia solution is 10-y
[OH-] of ammonia solution is 10-14/10-y = 10y-14
[H+] of hydrochloric acid solution is 10-x=10y-14
Is [H+] of ammonia solution not 10-x?
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Yea, sorry. I mixed up the x and y. Thanks
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Hint
strong acid, weak base
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[OH-] of ammonia solution is 10-14/10-y = 10y-14
[H+] of hydrochloric acid solution is 10-x=10y-14
I am sorry, but I don't understand these steps. Can someone explain this to me? Thanks.
pH of hydrochloric acid is y. Shouldn't the [H+] of hydrochloric acid solution be 10-y?
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[tex]pH + pOH = 14[/tex]
[tex][H^+] = 10^{-pH}[/tex]
[tex][OH^-] = 10^{-pOH} = 10^{-(14-pH)} = 10^{pH-14}[/tex]
Makes sense now?
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Hint
strong acid, weak base
Strong acids dissociate more than weak bases. I am finding it really hard to make the link. I think I am missing something very obvious. Going through the steps again:
1) Solution of ammonia has same amount of OH- as H+ in solution of HCl
2) When mixed, before any reacting occurs there is the same amount of OH-, H+, Cl- (since HCl is a strong acid), and ammonium. And there is a lot more ammonia since ammonia is a weak base.
3) The OH- and H+ react to produce water. Since the OH- has been removed, equilibrium makes the ammonia turn into ammonium. But since there is no more HCl (since it dissociated before the reacting) the OH- does not react and stays as OH-. So in the end:
There is a small amount of [H+] (left from the water production reaction)
More [Cl-]
And same amount of [NH4+] and [OH-] from the equilibrium shift in ammonia reaction
What am I missing? Thanks!
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I feel like the clue that the original pH of acid was <3 could be useful now. If you estimate the new pH of the mixed solution, you could then estimate ball park figures for the NH4+ and OH-
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[tex]pH + pOH = 14[/tex]
[tex][H^+] = 10^{-pH}[/tex]
[tex][OH^-] = 10^{-pOH} = 10^{-(14-pH)} = 10^{pH-14}[/tex]
Makes sense now?
I see, thanks.
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Strong acids dissociate more than weak bases
strong acid dissociate practically completely
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I feel like the clue that the original pH of acid was <3 could be useful now. If you estimate the new pH of the mixed solution, you could then estimate ball park figures for the NH4+ and OH-
Is it possible to calculate the pH of the mixed solution without the Ka and Kb? Because you need to account for the equilibrium shift in the NH3+H2O ::equil:: NH4++OH- reaction.
The only way I can solve this problem is through elimination. I know NH4+ will be the most concentrated so it leaves you with A and C. And I know OH- will be more than H+ because of the equilibrium shift so the answer is A. However, I don't know if [Cl-] > [OH-] or [OH-] > [Cl-], is it possible to do this without the Ka or Kb?
Thanks
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Okay, scratch the last suggestion. Try sketching bar graphs illustrating answers to the following 3 qus:
What was the relationship between the concentrations of the four species directly before mixing?
How did that relationship change directly after mixing?
How did it change having established equilibrium?
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What am I missing? Thanks!
And this reasoning is sufficient for choosing answer.
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Ok, thanks everyone for the *delete me*