iluvcats
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Can anyone help with these 2 questions please? The answer for Q12 is C and Q13 is D.

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Theo_Randall
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When more water is added this raises the pH as the conc of H+ decreases. HF is a weak acid so its Ka at a certain temp must stay constant. When all of the concentrations of the things in the Ka equation decrease, something must increase to keep Ka the same, in this case, more of the acid dissociates, increasing conc. of H+ and A-, so percentage dissociated must increase.
However, as there's no salt of the acid in solution, it can't act as a buffer, at least not very well, especially when volume is doubled, so conc. of H+ overall decreases, and you have a lower pH. Hence, C.

Ka= ([H+][A-])
[HA]

For the second question, your values of [A-] and [HA] are the same, so your equation ends up as Ka = [H+], and Ka is the same as they're the same acid, so pH is the sam for both. However, as there is more [HA] and [A-] in M, it requires a greater volume of acid to change its pH than N, so N will change first. Hence, D
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iluvcats
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(Original post by Theo_Randall)
When more water is added this raises the pH as the conc of H+ decreases. HF is a weak acid so its Ka at a certain temp must stay constant. When all of the concentrations of the things in the Ka equation decrease, something must increase to keep Ka the same, in this case, more of the acid dissociates, increasing conc. of H+ and A-, so percentage dissociated must increase.
However, as there's no salt of the acid in solution, it can't act as a buffer, at least not very well, especially when volume is doubled, so conc. of H+ overall decreases, and you have a lower pH. Hence, C.

Ka= ([H+][A-])
[HA]

For the second question, your values of [A-] and [HA] are the same, so your equation ends up as Ka = [H+], and Ka is the same as they're the same acid, so pH is the sam for both. However, as there is more [HA] and [A-] in M, it requires a greater volume of acid to change its pH than N, so N will change first. Hence, D
Thanks very much for your explanation! It makes things a lot clearer. I just have a question regarding pH during dissociation. It would make sense that the same mol of H+ is in a larger volume of water so pH decreases, but when you consider the equation for dilution:

HF + H2O --> H3O+ + F-

wouldn't you say that dilution would cause the equilibrium to shift to the right side and hence more H+ should be formed, thus decreasing pH?
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iluvcats
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In this question, the points in the mark scheme are:
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- CH3COOH + H2O --> CH3COO- + H3O+
- equilibrium shifts to the right so there's more dissociation
- [H+] is greater than expected

I'm assuming they are saying that because H+ is greater so the pH doesn't increase as much. However when comparing to a strong acid that completely dissociates, wouldn't the strong acid have an even higher concentration of H+ rather than a something that's partially dissociated?
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Theo_Randall
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(Original post by iluvcats)
Thanks very much for your explanation! It makes things a lot clearer. I just have a question regarding pH during dissociation. It would make sense that the same mol of H+ is in a larger volume of water so pH decreases, but when you consider the equation for dilution:

HF + H2O --> H3O+ + F-

wouldn't you say that dilution would cause the equilibrium to shift to the right side and hence more H+ should be formed, thus decreasing pH?
Yes, more of the acid dissociates because of Le Chatelier's principle, so there is more H+ however, this is outweighed by the effect of adding the water which decreases the conc. more than the dissociation increases it, so overall pH would increase.
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iluvcats
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(Original post by Theo_Randall)
Yes, more of the acid dissociates because of Le Chatelier's principle, so there is more H+ however, this is outweighed by the effect of adding the water which decreases the conc. more than the dissociation increases it, so overall pH would increase.
Ahh.. I see. Thank you! Any chance you know how to do that question from the second post too?
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(Original post by iluvcats)
Ahh.. I see. Thank you! Any chance you know how to do that question from the second post too?
Need to remember that a high H+ conc gives a LOW pH, and that diluting decreases H+ conc, and increases pH. Although diluted by factor 10, pH increases by less than one, as the dissociation is an equilibrium reaction. Standard Le Chatelier stuff, CH3COOH + H2O --> CH3COO- + H3O+. Son increasing H2O shifts equilibrium to right, which means more H+, but still diluted so pH still increases.

Hope that helps!
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iluvcats
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(Original post by Theo_Randall)
Need to remember that a high H+ conc gives a LOW pH, and that diluting decreases H+ conc, and increases pH. Although diluted by factor 10, pH increases by less than one, as the dissociation is an equilibrium reaction. Standard Le Chatelier stuff, CH3COOH + H2O --> CH3COO- + H3O+. Son increasing H2O shifts equilibrium to right, which means more H+, but still diluted so pH still increases.

Hope that helps!
I get that theory but why would the weak acid have more H+ compared to the strong acid? Say the strong acid had 1 mol/dm3 of H+ since it fully dissociates and the weak acid has 0.5 mol/dm3 of H+ since it partially dissociates. Dilution would cause an overall increase in pH, but wouldn't the strong acid hold back the increase in pH more than the weak acid does since the strong acid has a higher concentration of H+?
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