midterm question// Concentration ratio  [ENDORSED]

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kennedyp
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midterm question// Concentration ratio

Postby kennedyp » Fri Feb 14, 2020 8:56 am

I don't know if we all had the same midterm, but there was one question about concentrations and ph that I genuinely had no idea how to solve. It was about a molecule that would turn into a salt I think, and then i remember something about the ph of the stomach and the question asked us to find the ratio of the concentrations. How were we supposed to go about solving that problem?

Kaitlyn Ang 1J
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Re: midterm question// Concentration ratio  [ENDORSED]

Postby Kaitlyn Ang 1J » Fri Feb 14, 2020 9:35 am

What I did was I found Ka and used the pH given to find the concentration of H+. Since the equation for Ka is [H+][A-]/[HA], you can find the ratio of conjugate base to acid by dividing Ka by the concentration of H+.

Amy Xiao 1I
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Re: midterm question// Concentration ratio

Postby Amy Xiao 1I » Tue Feb 18, 2020 6:18 pm

Kaitlyn Ang 1J wrote:What I did was I found Ka and used the pH given to find the concentration of H+. Since the equation for Ka is [H+][A-]/[HA], you can find the ratio of conjugate base to acid by dividing Ka by the concentration of H+.



But how would you know if it stays as a salt in the stomach or becomes something else?

005321227
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Re: midterm question// Concentration ratio

Postby 005321227 » Wed Feb 19, 2020 1:30 am

was the equation given on the equation sheet on the midterm? I dont recall it being given

Catherine Daye 1L
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Re: midterm question// Concentration ratio

Postby Catherine Daye 1L » Wed Feb 19, 2020 1:34 am

You have to think about how Ka=[A][H+]/[CB] (CB being conjugate base). By rearranging the equation, you can get Ka/[H+] = [A]/[CB], which is a ratio.

Emily Lo 1J
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Re: midterm question// Concentration ratio

Postby Emily Lo 1J » Wed Feb 19, 2020 4:43 pm

Catherine Daye 1L wrote:You have to think about how Ka=[A][H+]/[CB] (CB being conjugate base). By rearranging the equation, you can get Ka/[H+] = [A]/[CB], which is a ratio.

Were we given this equation anywhere? I mean like the original, not manipulated, equation? And I don't recall using it before the midterm but I may be wrong.

Kaitlyn Ang 1J
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Re: midterm question// Concentration ratio

Postby Kaitlyn Ang 1J » Sun Feb 23, 2020 2:40 pm

Emily Lo 1J wrote:
Catherine Daye 1L wrote:You have to think about how Ka=[A][H+]/[CB] (CB being conjugate base). By rearranging the equation, you can get Ka/[H+] = [A]/[CB], which is a ratio.

Were we given this equation anywhere? I mean like the original, not manipulated, equation? And I don't recall using it before the midterm but I may be wrong.


We weren't given the equation explicitly, but anytime we calculate Ka, we use that equation implicitly. Like when we think of [products]/[reactants], the products are the conjugate base ([A-]) and the hydronium atoms [H3O+] and the reactants are the original acid ([HA]) and water, though water doesn't factor into the equation. So that's where the Ka = [H+][A-]/[HA] equation comes from.

Kaitlyn Ang 1J
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Re: midterm question// Concentration ratio

Postby Kaitlyn Ang 1J » Sun Feb 23, 2020 2:43 pm

Amy Xiao 1I wrote:
Kaitlyn Ang 1J wrote:What I did was I found Ka and used the pH given to find the concentration of H+. Since the equation for Ka is [H+][A-]/[HA], you can find the ratio of conjugate base to acid by dividing Ka by the concentration of H+.



But how would you know if it stays as a salt in the stomach or becomes something else?


The answer to the first part is (imo) kind of unrelated to the ratio question. To see if an acid is strong enough to protonate (aka gives off a H+), the pKa has to be smaller than the pH. Since in this case, the pKa is greater than the pH, the acid will not protonate, so it will not change forms.

705121606
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Re: midterm question// Concentration ratio

Postby 705121606 » Wed Feb 26, 2020 10:08 pm

Kaitlyn Ang 1J wrote:
Amy Xiao 1I wrote:
Kaitlyn Ang 1J wrote:What I did was I found Ka and used the pH given to find the concentration of H+. Since the equation for Ka is [H+][A-]/[HA], you can find the ratio of conjugate base to acid by dividing Ka by the concentration of H+.



But how would you know if it stays as a salt in the stomach or becomes something else?


The answer to the first part is (imo) kind of unrelated to the ratio question. To see if an acid is strong enough to protonate (aka gives off a H+), the pKa has to be smaller than the pH. Since in this case, the pKa is greater than the pH, the acid will not protonate, so it will not change forms.


What does it mean when pKa is smaller than pH? How does that tell us if it will protonate?

Kaitlyn Ang 1J
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Re: midterm question// Concentration ratio

Postby Kaitlyn Ang 1J » Sun Mar 08, 2020 11:59 pm

When pKa is smaller than pH, that means that Ka is larger than the H+ concentration, meaning that the products (H+ concentration included) is larger, so since there is a higher H+ concentration than the neutral pH indicates, this means that the acid protonates.


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