Thank you for this explanation. I actually found it much more easier to just do the eqn for #2 and 3. I get mixed up sometimes with conceptualizing contents.
You're welcome, and that's great! It's nice to have both the conceptual understanding and the equation in your 'toolbox' so you can handle any style of MCAT question on these topics!
Quick Review: pH=pKa+log(A/HA), pKa > pH = protonated & pKa < pH = deprotonated, base cxn is 10x the acid cxn => pH = pKa + 1 since more basic form exists when pKa < pH (100x => +2, 1000x => +3) , if pKa is pH + 1 => ratio of acid to base is 10:1 due to log scale. Question for Problem 1: I had assumed that when you meant neutral you meant an even ratio between acid and base; however, I did not critically think about the molecule shown. This led me to pick choice B, but you explained that neutral in this case would be a protonated form of OH since it would act as a leaving group. This would leave the molecule neutral. Is this reasoning correct and is there anything I should be thinking about? Huge thank you for the amazing videos Amanda. Have been watching a bunch lately despite me having to take the MCAT 5/24, wish I had found you sooner. Your teaching methods give a deeper understanding while maintaining an easy to understand flow, which is hard to find.
Another question: It is a bit hard for me to conceptualize the relationship between pKa, pH, and Ka. I understand the relationship between Ka and pH. A higher Ka means a higher dissociation of the acid, which is related to a lower pH or more acidic molecule. This means that molecule is more likely to produce H, donate H, or accept e- pair (Arrhenius, Bronsted, Lewis, respectively). Where does pKa fit into all this? Would love to hear your reasoning on this.
Yes - you've got it down! And yes, neutral is unfortunately a word that can mean several things in science! When we are describing molecules, neutral refers to uncharged. When we are referring to solutions, neutral refers to being at neutral pH. So glad you are finding the videos useful, and a month of quality studying can make a huge impact on your MCAT performance!
pKa is just taking the -log of ka (in fact any lower case p just means "power of" referring to a number that is the result of taking a logarithm of an exponent) So, just like pH = -log[H+], pka = -log(Ka). Because there's a negative sign in these equations, there's an inverse relationship between the values. A high Ka = more H+ = lower pH = lower pKa = more acidic. As scientists, we just use pKa and pH instead of Ka and [H+] because it's easier to deal with a positive linear scale (1, 2, 3, etc) than an exponential one (10^-1, 10^-2, 10^-3, etc). I have a video series that goes through the math related to these concepts here if you'd like to check it out! ua-cam.com/video/g3caAocfVUA/v-deo.htmlfeature=shared
@@bremmethod Thank you! This makes sense, dealing with a positive linear scale does make it easier, as long as you understand how it comes about. I'll check out the video and I'm looking forward to more!
When we look at functional groups in the context of Henderson-Hasselbalch, we first want to determine what is the protonated and deprotonated form. While alcohols can act as acids or bases, in this case our pKa of our molecule is 4, which means it is going to act as an acid and donate it's proton as soon as the pH gets above 4. For the hydroxy (OH) group, that means it will go from it's protonated form (-OH) to it's deprotonated form (O-). Remember that low pKas (7) will act as a B-L base and accept a proton, forming a conjugate acid (it's protonated form!) Hope this helps!
You are the best teacher I have ever learned from. Thank you for sharing this content. SO appreciated!
You are so welcome! I'm thrilled to hear that my lessons resonate with you!
Yooo were you was ?!! You came from heaven thank you very much ❤
haha thank you and I'm glad it was helpful!
Really being able to calculate it helped me understand better. Before I just memorized the rules but now I know how to calculate it. Thank you.
You're welcome! So glad it was helpful
You have helped so much. Thank uou
Glad the video helped!!
Thank you for this explanation. I actually found it much more easier to just do the eqn for #2 and 3. I get mixed up sometimes with conceptualizing contents.
You're welcome, and that's great! It's nice to have both the conceptual understanding and the equation in your 'toolbox' so you can handle any style of MCAT question on these topics!
this is soo helpful.
I'm so happy to hear that!
Quick Review: pH=pKa+log(A/HA), pKa > pH = protonated & pKa < pH = deprotonated, base cxn is 10x the acid cxn => pH = pKa + 1 since more basic form exists when pKa < pH (100x => +2, 1000x => +3) , if pKa is pH + 1 => ratio of acid to base is 10:1 due to log scale.
Question for Problem 1: I had assumed that when you meant neutral you meant an even ratio between acid and base; however, I did not critically think about the molecule shown. This led me to pick choice B, but you explained that neutral in this case would be a protonated form of OH since it would act as a leaving group. This would leave the molecule neutral. Is this reasoning correct and is there anything I should be thinking about?
Huge thank you for the amazing videos Amanda. Have been watching a bunch lately despite me having to take the MCAT 5/24, wish I had found you sooner. Your teaching methods give a deeper understanding while maintaining an easy to understand flow, which is hard to find.
Another question: It is a bit hard for me to conceptualize the relationship between pKa, pH, and Ka. I understand the relationship between Ka and pH. A higher Ka means a higher dissociation of the acid, which is related to a lower pH or more acidic molecule. This means that molecule is more likely to produce H, donate H, or accept e- pair (Arrhenius, Bronsted, Lewis, respectively). Where does pKa fit into all this? Would love to hear your reasoning on this.
Yes - you've got it down! And yes, neutral is unfortunately a word that can mean several things in science! When we are describing molecules, neutral refers to uncharged. When we are referring to solutions, neutral refers to being at neutral pH.
So glad you are finding the videos useful, and a month of quality studying can make a huge impact on your MCAT performance!
pKa is just taking the -log of ka (in fact any lower case p just means "power of" referring to a number that is the result of taking a logarithm of an exponent)
So, just like pH = -log[H+], pka = -log(Ka).
Because there's a negative sign in these equations, there's an inverse relationship between the values. A high Ka = more H+ = lower pH = lower pKa = more acidic.
As scientists, we just use pKa and pH instead of Ka and [H+] because it's easier to deal with a positive linear scale (1, 2, 3, etc) than an exponential one (10^-1, 10^-2, 10^-3, etc).
I have a video series that goes through the math related to these concepts here if you'd like to check it out! ua-cam.com/video/g3caAocfVUA/v-deo.htmlfeature=shared
@@bremmethod Thank you! This makes sense, dealing with a positive linear scale does make it easier, as long as you understand how it comes about. I'll check out the video and I'm looking forward to more!
if you can make a video on a short cut
for enthalpy solving for formation id appreciate it plz!
I'll add it to the list!
On problem 1, how did you know OH Wouk be negatively charged?
When we look at functional groups in the context of Henderson-Hasselbalch, we first want to determine what is the protonated and deprotonated form. While alcohols can act as acids or bases, in this case our pKa of our molecule is 4, which means it is going to act as an acid and donate it's proton as soon as the pH gets above 4. For the hydroxy (OH) group, that means it will go from it's protonated form (-OH) to it's deprotonated form (O-).
Remember that low pKas (7) will act as a B-L base and accept a proton, forming a conjugate acid (it's protonated form!)
Hope this helps!
@@bremmethodthank you
id like to see videos for cars
Check out this one - ua-cam.com/video/yt2Qd1v2gjw/v-deo.html