# Buffer Q (1 Viewer)

#### =)(=

##### Active Member

any clues on how to do this, the given solutions are straight up wrong

#### Masaken

##### Unknown Member
View attachment 41082
any clues on how to do this, the given solutions are straight up wrong
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

#### =)(=

##### Active Member

soz i dont but this is their sample ans

#### Masaken

##### Unknown Member
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]

#### 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

#### =)(=

##### Active Member
i got 6.09 but that is the method i applied.......

what paper is this from
parra 23

#### =)(=

##### Active 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]
cause like they say the concentration of h2co3 is going to be the old concentration + concentration of hcl which doesnt make sense

#### =)(=

##### Active 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]
also like they dont account for the shift of equilibrium like the just put the initial values in

#### Luukas.2

##### Active Member
also like they dont account for the shift of equilibrium like the just put the initial values in

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

$\bg_white K_{\text{a}} = \frac{x([\text{base}]+x)}{[\text{acid}]-x} = 4.30 \times 10^{-7}$

Make the usual simplifications and the value of x will come out at the answer they gave.

#### =)(=

##### Active Member

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

$\bg_white K_{\text{a}} = \frac{x([\text{base}]+x)}{[\text{acid}]-x} = 4.30 \times 10^{-7}$

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?

#### =)(=

##### Active Member

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

$\bg_white K_{\text{a}} = \frac{x([\text{base}]+x)}{[\text{acid}]-x} = 4.30 \times 10^{-7}$

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
$\bg_white K_{\text{a}} = \frac{x([\text{base}]-x)}{[\text{acid}]+x} = 4.30 \times 10^{-7}$

##### done hsc yay
also I might be wrong but should it be
$\bg_white K_{\text{a}} = \frac{x([\text{base}]-x)}{[\text{acid}]+x} = 4.30 \times 10^{-7}$
$\bg_white K_{\text{a}} = \frac{x([\text{base}]+x)}{[\text{acid}]-x} = 4.30 \times 10^{-7}$
Scroll up, that's what @Luukas.2 posted

OH SHIT MY BAD THEY'RE DIFFERENT SORRY LOL

#### carrotsss

##### New Member
also I might be wrong but should it be
$\bg_white K_{\text{a}} = \frac{x([\text{base}]-x)}{[\text{acid}]+x} = 4.30 \times 10^{-7}$
no, ur adding x to the RHS and removing from the LHS

#### =)(=

##### Active Member
ohh right ic so x is the change but doesnt that mean that you assume intial h3o is 0? soz if i am asking to many questions i am very tripped out by this