Bro, all I needed was this 2-minute video to explain this. I was looking for this for many hours as I have searched through complicated methods that somewhat explain what they do but isn't easy to track back. Thanks!
You are an undisputed genius. You deserve literally millions of subscribers. Thank you. This simple video saved me days of sifting through poorly explained lectures with no mechanical knowledge expanded on.
thanks alot, for some reson my teachers don't use this method and only teach the RICE chart. this saved me a lot of time on the test and study guide , keep being great.
For strong acids the molarity = the H+. Strong acids ionize close to 100%. So if you have a strong acid like HCl with a M of .8, the H+ = .8. Then you can just do the pH=-logH+ of the .8 to find pH. You don't need a Ka because the molarity = the H+ in a strong acid. This video link will show you this:ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-8MjesfHlWjQ.html
My online chem lecture spent 15 minuets solving this problem that ending in having to use the quadratic formula. Came across this video to find out I can solve the same question in basically 3 steps. THANK YOU THANK YOU THANK YOU.
This is a reasonable approximation, but it assumes the 5% rule. When the acid dissociates, the Ka value is only valid AT equilibrium. Consequently, the [HA] should be (0.45-x) because an x amount has already been lost to dissociation. This example answer is already rounded down slightly from the actual answer, but if Ka is higher, this method will slowly become less and less accurate. Any chemistry 2 teacher will likely point out that you can only use the 5% rule in certain cases, and this method is even more of a specific case on top of that.
I had figured it out. The reason is because I mistook the dot under the squared sign as multiplication, then I looked at the problem again and saw the dot was actually .45M. You have the dot written high up as multiplication instead of lower on the bottom of the number 4. Thanks for your time uploading this. You made it simple for me. Chemistry is frustrating to try to understand when you don't get it, but these videos are a real life saver for me.
Kw = Ka + Kb 1x10^-14 = Ka + Kb Kw always equals 1x10^-14 and youll have either Ka or Kb so subtract to find your unknown **edit: Kw equals 1x10^-14 only at 25 degrees Celcius but thats standard for these types of problems and should be assumed unless stated otherwide**
They may want you to look up Ka value in your textbook table but at minimum they should mention that you need to do that. Some Chem teachers make things harder than necessary.
