You are so good in explaining this! i understood this immediately! :)) may i request if you can make a slide version in PDF type or Power point for us to download ? :D it really helps since my textbook and lecture note is really hard to understand about this Buffer .
@FWGSD Yes, exactly. These form a conjugate acid-base pair, so a solution of both will be a buffer solution. If the concentrations are the same, then the pH of the buffer is equal to the pKa of the acid. In this case, pKa(NH4Cl) = 9.25 so this buffer is basic, with a pH of 9.25.
Let's say we have H2CO3 or H2SO4. Two ionization constants are given. They ask to calculate the pH. So, what constant should I use? Or should I use both of them and then sum two pH into one? It would be appreciated if You helped me out.
For any diprotic acid, the first ionization dominates and you can usually ignore the second ionization. For sulfuric acid, Ka1 is very large (strong acid) so the pH is determined approximately from the bulk acid: 0.2 M H2SO4 makes 0.2 M H+, so pH = -log(0.2)=0.67 and the second ionization contributes nothing. For H2CO3, Ka1 = 2.5E-4 (weak acid) so a 0.2M solution of carbonic acid gives [H+] = 0.00707 M or pH=2.15. The second ionization is MUCH weaker, so contributes almost nothing.
@Wrinkle5K In my CHM2046 class, the ka1 and ka2 are used to get their respective h30 concentrations, and then they are summed. Usually, the difference doesn't matter much though.
Great video, I've got the MCAT in two days and have always had trouble with buffers. I managed to formulate this stuff on my own on my last practice but it is good to finally understand it.
Sure! Suppose you have HCl (strong acid) and its salt, NaCl (chloride ion is an extremely weak base). The Cl- cannot compete for the H+ ions, so they remain in solution as H3O+ (strong acid). In terms of the math, the pH or solution is approximately equal to the pKa of the acid. If you have a strong acid like HCl, the Ka is about ~100, so the pKa is about 10^(-100) or nearly zero. That means you'll have a strongly acidic solution that essentailly cannot be buffered. OK?
chuck wight alriteeeeee
xLivexxLaughx 4 months ago
itz gud i learn so much frm here
verma352 6 months ago
@verma352 Try visitring some grammar and english language videos as well.
diamondboy9 2 months ago
You are so good in explaining this! i understood this immediately! :)) may i request if you can make a slide version in PDF type or Power point for us to download ? :D it really helps since my textbook and lecture note is really hard to understand about this Buffer .
yuukayuuto 8 months ago
This helped sooo much! Thank you! :)
Escadoo11 9 months ago
is a mixture of ammonia and ammonium chloride considered a buffer solution?
FWGSD 10 months ago
@FWGSD Yes, exactly. These form a conjugate acid-base pair, so a solution of both will be a buffer solution. If the concentrations are the same, then the pH of the buffer is equal to the pKa of the acid. In this case, pKa(NH4Cl) = 9.25 so this buffer is basic, with a pH of 9.25.
chemdog8 10 months ago
@chemdog8 Thanks for the help.
FWGSD 10 months ago
what does a conjugate acid/base actually mean?
armalik17 1 year ago
I think I will shoot myself in head!
WTF is he talking about?oO
Tell me how to cook amphetamines!!!
eSuMatix 1 year ago
thanks a lot for uploading this video...it really helps us to understand the buffer solution a bit more than reading in books
Scientist300 1 year ago
Hello. My question to You is:
Let's say we have H2CO3 or H2SO4. Two ionization constants are given. They ask to calculate the pH. So, what constant should I use? Or should I use both of them and then sum two pH into one? It would be appreciated if You helped me out.
Wrinkle5K 2 years ago
For any diprotic acid, the first ionization dominates and you can usually ignore the second ionization. For sulfuric acid, Ka1 is very large (strong acid) so the pH is determined approximately from the bulk acid: 0.2 M H2SO4 makes 0.2 M H+, so pH = -log(0.2)=0.67 and the second ionization contributes nothing. For H2CO3, Ka1 = 2.5E-4 (weak acid) so a 0.2M solution of carbonic acid gives [H+] = 0.00707 M or pH=2.15. The second ionization is MUCH weaker, so contributes almost nothing.
chemdog8 2 years ago
Thanks a lot. It's become much more clearly to me now.
Wrinkle5K 2 years ago
@Wrinkle5K In my CHM2046 class, the ka1 and ka2 are used to get their respective h30 concentrations, and then they are summed. Usually, the difference doesn't matter much though.
witeflight 1 year ago
I see. Thank You. =)
Wrinkle5K 1 year ago
thank you
this help me alot
lyac 2 years ago
Great video, I've got the MCAT in two days and have always had trouble with buffers. I managed to formulate this stuff on my own on my last practice but it is good to finally understand it.
lonelyjew 2 years ago
can i ask why for instance a strong acid and its salt cannot act as buffer solutions?
sedmercado 2 years ago
Sure! Suppose you have HCl (strong acid) and its salt, NaCl (chloride ion is an extremely weak base). The Cl- cannot compete for the H+ ions, so they remain in solution as H3O+ (strong acid). In terms of the math, the pH or solution is approximately equal to the pKa of the acid. If you have a strong acid like HCl, the Ka is about ~100, so the pKa is about 10^(-100) or nearly zero. That means you'll have a strongly acidic solution that essentailly cannot be buffered. OK?
chemdog8 2 years ago
complete ionization of the strong acid causes [H+] to increase therefore pH drastically drops...hope this helps
oridniv 2 years ago
absolutely great
this is so much help!
thank you!
kingcobrakhan 2 years ago