Quick Revision - Buffer solution calculations

[@matildawillcox1693]

Love that you include past questions! I tend to use Henderson-Hasselbalch but I think Kaacidoversalt has swayed me ahaha

[@MaChemGuy]

Matilda Willcox Haha indeed! Kaacidoversalt is the way forward. Good luck tomorrow 👍

[@matildawillcox1693]

@@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

[@MaChemGuy]

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

[@ammarafahad8094]

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?

[@Beautifulday1231r]

2:59 how did you know initial mol is 1

[@firminobecker7400]

Such a Time saver thankyou

[@MaChemGuy]

Firmino Becker Glad you approve!

[@firminobecker7400]

MaChemGuy would it be possible to make quick revision videos on as organic chem , like haloalkanes

[@MaChemGuy]

Firmino Becker Good shout. Will add to the list

[@MaChemGuy]

Firmino Becker Done!

[@firminobecker7400]

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 🙁

[@alexiaplacinta5468]

How do you know magnesium will react with hydrogen at 2:18 ?

[@scrumdum1]

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

[@nisargapoudel3911]

what if it reacted with the propanoic acid ? @@scrumdum1

[@tenfootman7600]

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 ?

[@chelsea9575]

How do you know the magnesium ions are going to react with the H+ ions?

[@MaChemGuy]

Chelsea Learned in Y12 acid + metal = salt + hydrogen and that it’s the H+ ion they react with

[@faizahbegum5257]

MaChemGuy what if they said a non metal?

[@mashaaln8764]

@@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.

[@mjian815]

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?

[@MaChemGuy]

Michael Jian That would be fine

[@DB-us6sb]

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

[@nosaii4660]

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?

[@malanhemal6574]

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?

[@teclar4457]

Can we use ka + log acid ÷salt

[@chlorine392]

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

[@mashalajan7419]

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]

Yup

[@babs8958]

why would it react with H+ not anything else

[@amaanali3414]

What if NaOH was in excess

[@magic2201]

Use kw to work out h+ ions the ph From there

[@ellaharwanko3497]

@@magic2201 would you use amount of moles left or the amount of moles reacted?

[@oswin4715]

Where does the 0.5 come from at 3:06?

[@MaChemGuy]

Oswin Mole ratio between Mg and H+

[@oswin4715]

@@MaChemGuy ahh thank you

[@hammadsyed1401]

At 8:02 you got 2.04 but how do you know that its a 1:2

[@MaChemGuy]

Because I calculated the acid : salt concentration ratio

[@hammadsyed1401]

​@@MaChemGuy oh so 10^-3.55 and 10^-3.86 is the ratio?

[@MaChemGuy]

@@hammadsyed1401 Correct. They sometimes ask ratio questions

[@hammadsyed1401]

@MaChemGuy thank you so much, your explanations are very clear.

[@MaChemGuy]

@@hammadsyed1401 you’re very welcome and thank you

[@annabelb4266]

at 8:18 why does the acid have to have twice the concentration of the salt

[@MaChemGuy]

Annabel B Because the ratio came out as 2:1

[@isabelokeke6255]

@@MaChemGuy but the equation is a 1:1 ratio

[@darkredsrevenge6154]

@@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

[@RukkaiahBae]

Awesome

[@j.2k561]

5:00 how come the mr of ch3coo- is 82

[@MaChemGuy]

Made from CH3COONa

[@RukkaiahBae]

Thx

[@CandyFace2000]

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.

[@MaChemGuy]

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

[@MJ177brrr]

2:38 y 2 moles of H+ Pls help I have my exams tomorrow

[@MaChemGuy]

Because Mg forms a 2+ ion

[@MJ177brrr]

@@MaChemGuy thank you so much for the reply ur explanations are so good.

[@MaChemGuy]

Good luck tomorrow!

[@MJ177brrr]

@@MaChemGuy thank you!