Mistake at 4:35. Multiplying the equal concentrations will actually result in a square, not necessarily a 2 times increase. This would also mean that dividing the dissociation constant of water by 2 would not give 10 to the -7. It would give 0.5x10^-14. Taking the square root of the constant would give the number you are looking for.
You are a saint. I'm taking my chemistry final and watched so many videos on how to calculate this and didn't understand any of them until I got to yours. Great video!
You are a savior. Tomorrow I will be taking my final for chemistry summer course and I was like planning skipping any question associated with these pH things. Now I am praying like, "Can all the questions be about pH?" Thank you.
Very helpful tips to use, and I like your formulas a lot better than the ones I got in my chemistry class. Thanks very much for explaining it thoroughly
This is fundamental to understand for chemists. For the Log, personally I simply enter the concentration and a negative number is computed, which I multiply mentally by -1. Then I have the pH (positive). This is quicker I think and less risk of error.
I’m reading the Campbell book on Biology. It says the electron of the hydrogen atom goes with one water molecule and the proton goes to the other water molecule. What happens to the rest of the hydrogen atom, specifically the neutron? Does it stay with the proton?
This video helped so much, but I still do not understand the way that very last problem was done. Working backwards with just the PH as the given.....I've been trying to figure it out for hours. Still clueless.
So basically 10^-pH will give you [H3O+] and 10^-pOH will give you the [OH-]. In this prompt, the pH (and thus the pOH) is given to us so we can work our way through it. The pH is 11.45, which means that 14-11.45=2.55 will be the pOH. All that is left is to plug this into your calculator and you should get easy results. Let me solve and make sure I gave the correct explanation. 10^-11.45 is equal to 3.5x10^-12 which is the [H3O+] and 10^-2.55 is 2.8x10^-3, which is what she found as a result. Hope this helped.
If in any doubt, just memorize them, but it's always best IMO to try to understand something than memorize it. So.... If you look at the 3 formulas at the end @22:50, the last 2 in green and pink are actually the same formula. Just wrote differently. So for example, assume the H30+ had a value of 0.316. 1) *green formula* pH=-Log( *0.316* ) => *0.5* 2) *pink formula* H30+ = 1 x 10 ^ -*0.5* => *0.316* The formulas in 1) and 2) are effectively the same thing. If you take the log of both sides of equation 1) you get equation 2). They've just been solved for you so you can just memorize them. One is the inverse of the other. Also, not wanting to confuse you but the pink formula gives you H30+. You can substitute this into the first Kw equation with effectively gives you.... OH- = 1.00 x 10 ^ *-14* / 1.00 x 10 ^ *-pH* When you have a power divided by a power you actually subtract them, so OH- = *14* - *pH*
![Calculating pH, pOH, [H+], [H3O+], [OH-] of Acids and Bases - Practice](http://i.ytimg.com/vi/UiK37I159fc/mqdefault.jpg)
Mistake at 4:35. Multiplying the equal concentrations will actually result in a square, not necessarily a 2 times increase. This would also mean that dividing the dissociation constant of water by 2 would not give 10 to the -7. It would give 0.5x10^-14. Taking the square root of the constant would give the number you are looking for.
x times x is x^2 ok
It does make sense if the concentrations were summed. Not sure if a product or sum is correct. More upset she didn't go into the logic of Kw
I really appreciate when she tells us which button to press coz I'm not good with a scientific calculator. Just bought new model Casio fx-570EX
x times x = x square and not 2x.
To solve for x, simply take the square root of Kw which will give 1 x e-7.
Thanks
Damar
You are a saint. I'm taking my chemistry final and watched so many videos on how to calculate this and didn't understand any of them until I got to yours. Great video!
You are a savior. Tomorrow I will be taking my final for chemistry summer course and I was like planning skipping any question associated with these pH things. Now I am praying like, "Can all the questions be about pH?" Thank you.
For ages, it's only now that I can say I understand how to do this calculations. Much love 💕😘
Thank you so much 💗💗💗💗💗💗🥰
Very helpful tips to use, and I like your formulas a lot better than the ones I got in my chemistry class. Thanks very much for explaining it thoroughly
YOU have earned a spot at the top of word document on chemistry calculations how to link. Thank You so much for this video!
I have a lab with this tomorrow and this is incredibly helpful for my pre-lab questions. Thank you so much for being so detailed and clear!
I’m glad I found this video it’s Clear and simple…ways to solve. Thank you!
saved my lifeee i spent foreverrr looking for some type of help lifesaverrrrrrrrr
God bless you! this is the first chemistry math video I have clearly understood and my final is in 3 days.
Good luck on your final!!
This helped ALOT, Thank you so much! Well explained and talked through!
This is fundamental to understand for chemists. For the Log, personally I simply enter the concentration and a negative number is computed, which I multiply mentally by -1. Then I have the pH (positive). This is quicker I think and less risk of error.
great suggestion!!
You are an amazing teacher!🥇❤
The explanation is explicit 'i love it.
Really well thought out explanation. Thank you!
Thank you so much for breaking this down and noting how errors happen.
omg thank you for this video!!!!! you saved my life
thank you so much for the calculator tip i was so lost doing my practice problems
I love reversible reactions, i also love le chatelier’s principle and henderson and hasselbach they are my friends for life😊
Bless you, woman, for these videos!
Thank you! Very clear explanation
Even though I saw your mistake, you have it all explained clearly. It would have been perfect but sill I thank you.
Very very helpful and well explained!
solid tutorial. chem instructor at Savannah Tech owes you part of her salary.
Interesting, simplified
I’m reading the Campbell book on Biology. It says the electron of the hydrogen atom goes with one water molecule and the proton goes to the other water molecule. What happens to the rest of the hydrogen atom, specifically the neutron? Does it stay with the proton?
excellent tutorial! new subscriber :)
Why calculate the pH And poH of a solution
This video helped so much, but I still do not understand the way that very last problem was done. Working backwards with just the PH as the given.....I've been trying to figure it out for hours. Still clueless.
So basically 10^-pH will give you [H3O+] and 10^-pOH will give you the [OH-]. In this prompt, the pH (and thus the pOH) is given to us so we can work our way through it. The pH is 11.45, which means that 14-11.45=2.55 will be the pOH. All that is left is to plug this into your calculator and you should get easy results. Let me solve and make sure I gave the correct explanation. 10^-11.45 is equal to 3.5x10^-12 which is the [H3O+] and 10^-2.55 is 2.8x10^-3, which is what she found as a result. Hope this helped.
If in any doubt, just memorize them, but it's always best IMO to try to understand something than memorize it. So....
If you look at the 3 formulas at the end @22:50, the last 2 in green and pink are actually the same formula. Just wrote differently.
So for example, assume the H30+ had a value of 0.316.
1) *green formula* pH=-Log( *0.316* ) => *0.5*
2) *pink formula* H30+ = 1 x 10 ^ -*0.5* => *0.316*
The formulas in 1) and 2) are effectively the same thing. If you take the log of both sides of equation 1) you get equation 2).
They've just been solved for you so you can just memorize them. One is the inverse of the other.
Also, not wanting to confuse you but the pink formula gives you H30+. You can substitute this into the first Kw equation with effectively gives you....
OH- = 1.00 x 10 ^ *-14* / 1.00 x 10 ^ *-pH*
When you have a power divided by a power you actually subtract them, so OH- = *14* - *pH*
Thank you for this video studying for my chem final
Thank you😻😻
Thanks maadam for your assistance
Thank you
Ty soòooo much yaaaaaaaayyyy 😊😊
Thank you for the tip about the calculator!😭
Hi
Thank you T.T
Ohhh I did a mistake and forgot to put a ( ) around it on the calculator 😅😂
This is so helpful, thanks
At 22:56 my calculator gives me an error for that pH calculation. I'm lost. But thanks for your time spent.
Ty soo much 😊
thank you so much, you came in clutch
Thank you! You saved me.
Subbed 👍🏽
10:49 another note, should be -5, not -6
If pH is starting 1×10 does not that mean my pH is always to the power? Thank you
5:00 Wrong! xx=x^2 (x to the second power, then getting the square root of both side will give you 1.00ox10-7, elementary middle school math!
Omg thank youuuuuuuuuu!!!!!!
oh my God you so awesome
Thank you very much
How did you get the poh as 2.55?
Because pH+pOH=14
HahHhaha you forgot it like me you must put it in ( ) in the calculator.
Try it
@@sanjitdas2038 that works too
how do u multiple a decimal with a sci notation?
like when solving OH- using the reaction of the chemicals where x2 = Ka/ molarity of the compound
And............ I'm saved.
x times x is x^2 ok
(x)(x)=x²....not 2x
Here you can post your questions and get quick answers or search for existing questions on the website.
starofgeeks.com
poor explaination