yes and no.
pH = pKa + log (base/acid)
You are right that base is acetate and acid is acetic acid.
You know pH = -log (H^+)
pKa is done the same way.
pKa = -log Ka. If you use 1.8 x 10^-5 for Ka of acetic acid, pKa = 4.74.
Let's call acetic acid HAc, sodium acetate is NaAc and acetate ion is Ac^-.
(HAc) = 0.1 M x (20 mL/45 mL) = ?? in the original solution.
(Ac^-) = 0.1 M x (25 mL/45 mL) = ?? in the original solution.
When adding HCl, the Ac^- ties it up as HAc (which ionizes only partially and that's why this is a good buffer--it won't let the HCl loose to ionize completely) so HAc increases by the amount of HCl added and the Ac^- decreases by the amount of HCl added. My advice is to work in moles or millimoles instead of concn It makes it simpler.
IN the polystyrene beaker, mix 20mL of 0.1 M Acetic acid and 25mL of 0.1M Sodium Acetate and immediately measure the pH. Remove the electrode and add 5ml of 0.1 M HCl to this buffer. Stir the solution and measure the pH.
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INFO:
concentration of acetic acid= 0.1M- 20ml
concentration sodium acetate= 0.1M- 25ml
concentration of hydrochloric acid- 0.1M - 5ml
pH measured:
Buffer solution- 4.69
Buffer + 5ml 0.1M HCL- 4.43
pH calculated????????????
Buffer solution- ?????
Buffer + 5ml 0.1M HCL- ?????
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PART A:
1.So i have to calculate pH before the addition of HCL
2. show CALCULATIONS for calculated pH after addition of HCl
I think i use the henderson hasselbalch equation:
pH= pka+log [A-]/[HA]
pH= pka+log [0.1][0.1]
I dunt know if i plugged in the right stuff and how do i calculate pka?????
1 answer
1 answer