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 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.
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!
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*
![pH Calculations - Calculate [H3O+] and [OH-], and Find the pH of a Solution](/img/logo.svg){kind=logo}
