Profesor Dave, I've used the formula pH= 1/2(pka - log(acide concentration) and solved for pka, I've got pka = 2pH + log(acide concentration) and then to get the ka I've just raised 10 to the negative pka, but I've got ka= 4.05*10^-4 instead of 4.5*10^-4 and I can't find out where I made a mistake, can you explaine please
I'm not sure what the mistake is but that isn't the equation I recommend you use to solve the problem. take 10^-pH to find the hydronium molarity which will be the same as the conjugate base molarity. Then divide it by the quantity of the initial acid molarity - the hydronium molarity. Now I see this was from two years ago. rip. oh well
@@seikomyazawa lol just wanted to ask the exact same question as you! I also got ka= 4.05*10^-4 initially, but then used the proper full formula K=(C*alpha^2 ) / 1-alpha and got 4,4425586*10^-4. (for alpha >5% full formula is a must) Mine result is more accurate than professor's Dave - exact even, worked on proper calculator without any approximations
YESSS! i also got this answer by using the initial concentration of the HNO2 as 0.0516 instead of using the equilibrium concentration which in the video he says is 0.047. I think the answers are pretty similar so it should be good enough?
How do I calculate both concentration and pH of all molecules and ions, if I'm only given that a propinoic acid in an aqueous solution has a Ka = 1,3*10^-5 and a concention on 0,100M ? I've made an ice box, but the process kind of stops then..
You probably dont care at all but does someone know of a method to get back into an instagram account..? I was dumb lost the password. I appreciate any tricks you can give me!
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Hey, I need to calculate how much the quantity of Conc H2SO4 Will be required for a 100 mL solution to reduce the pH from 6.5 to 4.5. How do I calculate it ?? Thank you so much in advance
