Achieve Question 13
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Achieve Question 13
Unsure this is the correct location to ask, but I am confused on question 13 on how you determine if the acid is charged or neutral! Any advice would help :)
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Re: Achieve Question 13
Hi!
When the pH of the solution is equal to the pKa (acid dissociation constant, if increased, acid is weaker) of the weak acid, there are the same amount of the neutral(HA) and charged species (A-) in the solution. This means the overall species is neutral. If the pH is less than the pKa, then the solution has mostly neutral species (HA) and is then more acidic. If the pH is greater than the pKa, however, this means that the acid has dissolved more in the solution and there are more charged A- species so the solution is more basic (in this case because A- is an anion base). This would mean that the predominant species was charged.
This is what I understood from the solution on achieve but I'm not sure how much of it makes sense. Hope this helps!
When the pH of the solution is equal to the pKa (acid dissociation constant, if increased, acid is weaker) of the weak acid, there are the same amount of the neutral(HA) and charged species (A-) in the solution. This means the overall species is neutral. If the pH is less than the pKa, then the solution has mostly neutral species (HA) and is then more acidic. If the pH is greater than the pKa, however, this means that the acid has dissolved more in the solution and there are more charged A- species so the solution is more basic (in this case because A- is an anion base). This would mean that the predominant species was charged.
This is what I understood from the solution on achieve but I'm not sure how much of it makes sense. Hope this helps!
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Re: Achieve Question 13
The way that I think about it is that if the pH of the solution is higher than the pKa of the acid, then the acid is more acidic then the solution and thus the forward reaction of the acid will occur and produce A- and H+. However, if the pH of the solution is lower than the pKa of the acid, the solution will be more acidic than the acid and thus the backwards reaction will be favored in which the solution will be deprotonated instead of the acid and it will form HA, the neutral form of the acid. Hope this helps clear things up a bit more!
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Re: Achieve Question 13
Hi! To answer the problem 13:
if pH > pKA : the solution is more basic, meaning it would go toward A- (the base) ; this means it would be negatively charged
if pH < pKA : the solution is more acidic, meaning it would go toward HA (the conjugate acid); this means it would be neutral since HA has no charge.
Generally speaking, so you can apply this to future problems:
if pH > pKA : the solution is more basic, meaning it would go toward the (sometimes conjugate) base ; the solution's charge is based on that compound's charge. Referring back to problem 13, A- was base, so if pH > pKA, the solution would've been negatively charged (A- is negatively charged)
if pH < pKA : the solution is more acidic, meaning it would go toward the (sometimes conjugate) acid ; the solution's charge is based on that compound's charge. Using an example from problem 14, B(aq)+H2O(l) ⇌ BH+(aq)+OH−(aq), if pH < pKA, we would refer to the conjugate acid BH+, meaning that the solution would be charged.
Hope this helped!
if pH > pKA : the solution is more basic, meaning it would go toward A- (the base) ; this means it would be negatively charged
if pH < pKA : the solution is more acidic, meaning it would go toward HA (the conjugate acid); this means it would be neutral since HA has no charge.
Generally speaking, so you can apply this to future problems:
if pH > pKA : the solution is more basic, meaning it would go toward the (sometimes conjugate) base ; the solution's charge is based on that compound's charge. Referring back to problem 13, A- was base, so if pH > pKA, the solution would've been negatively charged (A- is negatively charged)
if pH < pKA : the solution is more acidic, meaning it would go toward the (sometimes conjugate) acid ; the solution's charge is based on that compound's charge. Using an example from problem 14, B(aq)+H2O(l) ⇌ BH+(aq)+OH−(aq), if pH < pKA, we would refer to the conjugate acid BH+, meaning that the solution would be charged.
Hope this helped!
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