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Buffer Q (1 Viewer)
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Masaken
Unknown Member
do you have the correct answers at least? i have an answer but i just want to confirm if it's correct
you would definitely start by applying henderson-hasselbalch equation though
SB257426
Very Important User
Masaken
Unknown Member
just curious why it's wrong? this is how you would do it, i got the exact same answer i didn't do the exact way they did it [i did henderson-hasselbalch, but i guess the logic they use without the eq is the same as using the eq, there was a similar q in my trial and you could do it this way or apply the eq and you'd get the full marks]
Masaken
Unknown Member
how i did it:
when you add HCl, that's 0.015 moles to 1L of buffer. since you're ignoring any change in total volume and HCl ionises completely, [HCl] = [H+] = 0.015mol/L. so, applying henderson-hasselbalch to get the pH and subbing in the provided values given to you, that's pH = -log(4.3 x 10^-7) + log[(0.14 - 0.015)/(0.25 + 0.015]). adding H+ will take away from the base, and add to the acid. you get 6.040195...
before adding HCl, the pH of the buffer solution was 6.1129..., aligning with how buffers resist changes in pH
when you add HCl, that's 0.015 moles to 1L of buffer. since you're ignoring any change in total volume and HCl ionises completely, [HCl] = [H+] = 0.015mol/L. so, applying henderson-hasselbalch to get the pH and subbing in the provided values given to you, that's pH = -log(4.3 x 10^-7) + log[(0.14 - 0.015)/(0.25 + 0.015]). adding H+ will take away from the base, and add to the acid. you get 6.040195...
before adding HCl, the pH of the buffer solution was 6.1129..., aligning with how buffers resist changes in pH
parra 23
cause like they say the concentration of h2co3 is going to be the old concentration + concentration of hcl which doesnt make sense
also like they dont account for the shift of equilibrium like the just put the initial values in
The answer is right.
The working is poor.
If you set up an ICE table with the initial moles of the acid and conjugate base forms, then consider that the added HCl has the effect of converting the base form into the acid - so the acid moles goes up by n(H+) added and the base form goes down by the same amount, then convert the resulting values to concentrations using the new total volume, then figure the system moves right (to produce some H+, call that x, then you get an equation like
Make the usual simplifications and the value of x will come out at the answer they gave.
ohhh i get what ur saying but would it be wrong like to make an ice table and put initial concentration of h30+=[HCO3-]+ 0.14 and going from there cause like before adding hcl their concentrations are equal right?The answer is right.
The working is poor.
If you set up an ICE table with the initial moles of the acid and conjugate base forms, then consider that the added HCl has the effect of converting the base form into the acid - so the acid moles goes up by n(H+) added and the base form goes down by the same amount, then convert the resulting values to concentrations using the new total volume, then figure the system moves right (to produce some H+, call that x, then you get an equation like
Make the usual simplifications and the value of x will come out at the answer they gave.
also I might be wrong but should it beThe answer is right.
The working is poor.
If you set up an ICE table with the initial moles of the acid and conjugate base forms, then consider that the added HCl has the effect of converting the base form into the acid - so the acid moles goes up by n(H+) added and the base form goes down by the same amount, then convert the resulting values to concentrations using the new total volume, then figure the system moves right (to produce some H+, call that x, then you get an equation like
Make the usual simplifications and the value of x will come out at the answer they gave.
also I might be wrong but should it be
}{[\text{acid}]+x} = 4.30 \times 10^{-7})
Scroll up, that's what @Luukas.2 posted
OH SHIT MY BAD THEY'RE DIFFERENT SORRY LOL
lol
no, ur adding x to the RHS and removing from the LHSalso I might be wrong but should it be