Calculate the pH, pOH, and percentage deprotonation of each of the following aqueous solutions:
a) 0.20M CH3COOH(aq)
Can someone help me out please?
12.55
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Ben Rolnik 1D
- Posts: 33
- Joined: Fri Jun 23, 2017 11:39 am
Re: 12.55
yes
For this equation:
pH = -log( square root ( kA * molarity HA ).
That's not always true, but it works here... and also usually where you're not given much more info. This works because from your icebox the equation can be rounded to:
kA = X^2 / [Molarity HA]
////
pOH is always "14 - pH" because pH + pOH = 14.
Percentage deprotination is
[ Molarity H+ ] / [Initial Molarity]
You can think of this like the Percent Yield we did earlier in the course. It essentially means how much of your starting HA actually turned into H... in this case when you solve for pH using the equation above, you must take 10^(-pH) and that is your [H] Molarity....
Hope this helps!
For this equation:
pH = -log( square root ( kA * molarity HA ).
That's not always true, but it works here... and also usually where you're not given much more info. This works because from your icebox the equation can be rounded to:
kA = X^2 / [Molarity HA]
////
pOH is always "14 - pH" because pH + pOH = 14.
Percentage deprotination is
[ Molarity H+ ] / [Initial Molarity]
You can think of this like the Percent Yield we did earlier in the course. It essentially means how much of your starting HA actually turned into H... in this case when you solve for pH using the equation above, you must take 10^(-pH) and that is your [H] Molarity....
Hope this helps!
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