do you have the correct answers at least? i have an answer but i just want to confirm if it's correctView attachment 41082
any clues on how to do this, the given solutions are straight up wrong
i got 6.09 but that is the method i applied.......View attachment 41086
soz i dont but this is their sample ans
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]View attachment 41086
soz i dont but this is their sample ans
parra 23i got 6.09 but that is the method i applied.......
what paper is this from
cause like they say the concentration of h2co3 is going to be the old concentration + concentration of hcl which doesnt make sensejust 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]
also like they dont account for the shift of equilibrium like the just put the initial values injust 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]
The answer is right.also like they dont account for the shift of equilibrium like the just put the initial values in
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
Scroll up, that's what @Luukas.2 posted
lol
no, ur adding x to the RHS and removing from the LHSalso I might be wrong but should it be