Walkthroughs of four different buffer solution past exam questions. Each one tests a different aspect of the topic. A real test of your buffer solution understanding!
for the last question, I used HA=H+squared/ka to get a value for the weak acid and then rearranged the kaacidoversalt to get concentration of salt. I got 0.000576moldm3 for acid and 0.000282moldm3 for salt which is pretty much the 2.04/1 ratio of acid/salt. would this get me any marks??
@@MaChemGuy Hi, BIG fan here. I'm confused, I'm pretty sure the Mg doesn't react with H+ ions (since Mg isn't negatively charged) Question 1 Part 2 Correct me if I'm wrong, but I think its the C2H5COOH reacting with the Mg that makes (C2H5COO)2Mg and that then FULLY dissociates into 2C2H5COO- and Mg+ (because aq) so hence you get a reduction of 0.5 mol of C2H5COOH but an increase of 0.5 mol of C2H5COO- ---- p.s I came to this conclusion from watching your other vids.
It is because group 2 elements react with acids to produce salts and hydrogen gas, they donate the outer 2 electrons to the hydrogen ions in the acid so form Mg2+ + H2 So I think it doesn't react with the H+ ions in the right side of the buffer equation but the hydrogen in the actual acid in the left. Then you'll get 2C2H5COOH + Mg > (C2H5COO)2Mg + H2, which shows why the moles of acid drops by 0.5 and the moles of C2H5COO- increases by 0.5
Could you have also rearranged it to get Ka/[H+]= [salt]/[acid] because when I tried it I did that and got 0.49moldm13 of salt over 1moldm-3 acid which is pretty much the same ratio? Thank you.
That's fine, just be careful you get the ratio the right way round when giving your answer. The way I rearranged gives the ratio [Acid]/[Salt] ratio directly (as asked in the Q) whereas your way gives it the other way round. Hope that helps
@@isabelokeke6255 Acid/Salt = 10^-3.55/10^-3.86. Look at the numbers from the inverse log on your calculator. Acid conc is around double the conc of the salt from this, which is why the value came out as 2.04
![[ไฮไลต์] ฟุตบอลชาย อาร์เจนติน่า vs โมร็อกโก รอบแบ่งกลุ่ม | โอลิมปิก 2024](http://i.ytimg.com/vi/wE6uYH0yyzM/mqdefault.jpg)
Love that you include past questions! I tend to use Henderson-Hasselbalch but I think Kaacidoversalt has swayed me ahaha
Matilda Willcox Haha indeed! Kaacidoversalt is the way forward. Good luck tomorrow 👍
@@MaChemGuy Thank you! I had a bit of a meltdown earlier and decided to watch all your vids and they've helped calm me down a bit x
Matilda Willcox That’ll be my monotonous voice (as someone said I had today) Good that you’re all calm now, try and stay that way
How comes you added 0.5 moles to the final products? I thought that on addition of H+ ions the equilibrium shifted to the left
at 3:48 sir, do you need to know the molar ratio from before, for Mg reaction with H+ that 2H+ reacts with Mg, inorder to do the question?
Hi Sir, at 2:28 could u plz explain again why the buffer moves to the right and why does the acid conc goes down and salt conc goes up?
Machemguy @ 3:35 let's say an acid was added instead of Mg would the acid final moles be 1.5 and the salt final moles be the same so 1 ?
Such a Time saver thankyou
Firmino Becker Glad you approve!
MaChemGuy would it be possible to make quick revision videos on as organic chem , like haloalkanes
Firmino Becker Good shout. Will add to the list
Firmino Becker Done!
MaChemGuy just checked it out thank-you so much , my only issue now is trying to retain the entire spec and being equally good at AS as A2 🙁
at 8:22 can we use the concentrations as 2.04 moldm-3 acid and 1.00 moldm-3 salt as this is the exact ratio given from the calculation?
Michael Jian That would be fine
Can we use ka + log acid ÷salt
at 3:15 why are the final moles 0.5 for HA and 1.5 for A- if 0.5 moles of H+ is removed?
for the last question, I used HA=H+squared/ka to get a value for the weak acid and then rearranged the kaacidoversalt to get concentration of salt. I got 0.000576moldm3 for acid and 0.000282moldm3 for salt which is pretty much the 2.04/1 ratio of acid/salt. would this get me any marks??
Yup
How do you know the magnesium ions are going to react with the H+ ions?
Chelsea Learned in Y12 acid + metal = salt + hydrogen and that it’s the H+ ion they react with
MaChemGuy what if they said a non metal?
@@MaChemGuy Hi, BIG fan here.
I'm confused, I'm pretty sure the Mg doesn't react with H+ ions (since Mg isn't negatively charged)
Question 1 Part 2
Correct me if I'm wrong, but I think its the C2H5COOH reacting with the Mg that makes (C2H5COO)2Mg and that then FULLY dissociates into 2C2H5COO- and Mg+ (because aq) so hence you get a reduction of 0.5 mol of C2H5COOH but an increase of 0.5 mol of C2H5COO- ---- p.s I came to this conclusion from watching your other vids.
How do you know magnesium will react with hydrogen at 2:18 ?
It is because group 2 elements react with acids to produce salts and hydrogen gas, they donate the outer 2 electrons to the hydrogen ions in the acid so form Mg2+ + H2
So I think it doesn't react with the H+ ions in the right side of the buffer equation but the hydrogen in the actual acid in the left.
Then you'll get 2C2H5COOH + Mg > (C2H5COO)2Mg + H2, which shows why the moles of acid drops by 0.5 and the moles of C2H5COO- increases by 0.5
what if it reacted with the propanoic acid ? @@scrumdum1
Awesome
Thx
Where does the 0.5 come from at 3:06?
Oswin Mole ratio between Mg and H+
@@MaChemGuy ahh thank you
in min 3:46 you put salt over acid cux acid is 1.5 not 0.5 the salt in 0.5 or am i wrong
Could you have also rearranged it to get Ka/[H+]= [salt]/[acid] because when I tried it I did that and got 0.49moldm13 of salt over 1moldm-3 acid which is pretty much the same ratio? Thank you.
That's fine, just be careful you get the ratio the right way round when giving your answer. The way I rearranged gives the ratio [Acid]/[Salt] ratio directly (as asked in the Q) whereas your way gives it the other way round. Hope that helps
why would it react with H+ not anything else
at 8:18 why does the acid have to have twice the concentration of the salt
Annabel B Because the ratio came out as 2:1
@@MaChemGuy but the equation is a 1:1 ratio
@@isabelokeke6255 Acid/Salt = 10^-3.55/10^-3.86. Look at the numbers from the inverse log on your calculator. Acid conc is around double the conc of the salt from this, which is why the value came out as 2.04
What if NaOH was in excess
Use kw to work out h+ ions the ph From there
@@magic2201 would you use amount of moles left or the amount of moles reacted?
At 8:02 you got 2.04 but how do you know that its a 1:2
Because I calculated the acid : salt concentration ratio
@@MaChemGuy oh so 10^-3.55 and 10^-3.86 is the ratio?
@@hammadsyed1401 Correct. They sometimes ask ratio questions
@MaChemGuy thank you so much, your explanations are very clear.
@@hammadsyed1401 you’re very welcome and thank you
5:00 how come the mr of ch3coo- is 82
Made from CH3COONa
2:38 y 2 moles of H+
Pls help I have my exams tomorrow
Because Mg forms a 2+ ion
@@MaChemGuy thank you so much for the reply ur explanations are so good.
Good luck tomorrow!
@@MaChemGuy thank you!