HSC OVERALL MARKS GOAL AND/OR REALISTIC PREDICTIONS (1 Viewer)

Luukas.2 · Oct 6, 2023

$n(\text{HA}) = CV = 0.055 \times 45.00 \times 10^{-3} = 2.475 \times 10^{-3}\ \text{mol}$

$n(\text{NaOH}) = CV = 0.060 \times 20.00 \times 10^{-3} = 1.20 \times 10^{-3}\ \text{mol}$

So, once all the NaOH is consumed in the 1:1 reaction of NaOH and HA, the mixture consists of:

$n(\text{A}^-) = 1.20 \times 10^{-3}\ \text{mol}$
$n(\text{HA}) = (2.475 - 1.20) \times 10^{-3} = 1.275 \times 10^{-3}\ \text{mol}$ .

plus sodium ions in an aqueous solution. These can be converted to concentrations using the total volume of 65.00 mL and put into an ICE table, yielding

$K_{\text{a}} = \cfrac{\left[\text{H}_3\text{O}^+\right]\left[\text{A}^-\right]}{[\text{HA}]} = \cfrac{x(0.01846... + x)}{0.01961... - x} = 1.585 \times 10^{-9}$

where $x$ is the concentration of hydronium ions at equilibrium. Solving this with the usual assumptions - that $x << 0.01846...$ - will give:

$\begin{align*} \left[\text{H}_3\text{O}^+\right] &= x = \cfrac{K_{\text{a}}(0.01961... - x)}{0.01846... + x} \\ &\approx \cfrac{1.585 \times 10^{-9} \times 0.01961...}{0.01846...} \\ &= 1.6840... \times 10^{-9}\ \text{mol L}^{-1} \\ \\ \text{Thus,} \quad \text{pH} &= -\log_{10}\left(1.6840... \times 10^{-9}\right) \\ &\approx 8.774 \end{align*}$

The same result will be obtained using the Henderson-Hasselbalch equation:

$\begin{align*} \text{pH} &= \text{p}K_{\text{a}} + \log_{10}{\cfrac{\left[\text{A}^-\right]}{\left[\text{HA}\right]}} \\ &= -\log_{10}{\left(1.585 \times 10^{-9}\right)} + \log_{10}{\cfrac{0.01846...}{0.01961...}} \\ &\approx 8.774 \end{align*}$

And there is only one small problem with this calculation and answer... it is clearly wrong! If I take a weak acid solution and neutralise less than half of it, I can't possibly get a basic solution.

So, where is the problem?

The Henderson-Hasselbalch solution is incorrect as it is derived for a system at equilibrium, and so with the concentrations of the acidic and basic forms being equilibrium concentrations. More problematically, the ICE table has started with a solution of HA and A^- and the hydronium concentration being zero. This is impossible in any aqueous system, but is usually not a problem as the contribution of the equilibrium system to the hydronium concentration is much greater than is the contribution from the auto-ionisation of the water / solvent. If you look at the calculated hydronium concentration above, you can see that it is smaller than than the amount in neutral water at room temperature. So, to find the actual pH, a quick-and-dirty solution is to add in the contribution from the solvent to the contribution from the equilibrium:

$\begin{align*} \left[\text{H}_3\text{O}^+\right]_{eqm} &\approx 10^{-7} + x \\ &= 10^{-7} + 1.6840... \times 10^{-9} \\ &= 1.016840... \times 10^{-7}\ \text{mol L}^{-1} \\ \\ \text{Thus,} \quad \text{pH} &= -\log_{10}\left(1.016840... \times 10^{-7}\right) \\ &\approx 6.993 \end{align*}$

I don't think figuring this out is a reasonable expectation at HSC level.

itsalovehaterelo · Oct 6, 2023

ExtremelyBoredUser said:
idk im just giving realistic predictions idk why everyones getting mad like im being deadass no joke

nobody fr prepares eco essays tho...

Luukas.2 · Oct 6, 2023

Unovan said:
Ka and pka for strong acid is redundant in hsc bc its assumed strong acids ionise completely hence ka = infinity

That's true, for strong acids using pK_a values is largely irrelevant for the HSC... my point was that the question states that perchloric acid has a pK_a near 9 when it is actual near -15... that's out by 24 orders of magnitude or a factor of 100 million cubed. That's a lot!

Unovan · Oct 6, 2023

Luukas.2 said:
That's true, for strong acids using pK_a values is largely irrelevant for the HSC... my point was that the question states that perchloric acid has a pK_a near 9 when it is actual near -15... that's out by 24 orders of magnitude or a factor of 100 million cubed. That's a lot!

yeh I assume they just made a weak acid neut question, probably using another paper question as inspiration and subbed in a random acid without checking its strength

ExtremelyBoredUser · Oct 6, 2023

Luukas.2 said:
$n(\text{HA}) = CV = 0.055 \times 45.00 \times 10^{-3} = 2.475 \times 10^{-3}\ \text{mol}$

$n(\text{NaOH}) = CV = 0.060 \times 20.00 \times 10^{-3} = 1.20 \times 10^{-3}\ \text{mol}$

So, once all the NaOH is consumed in the 1:1 reaction of NaOH and HA, the mixture consists of:

$n(\text{A}^-) = 1.20 \times 10^{-3}\ \text{mol}$

$n(\text{HA}) = (2.475 - 1.20) \times 10^{-3} = 1.275 \times 10^{-3}\ \text{mol}$ .

plus sodium ions in an aqueous solution. These can be converted to concentrations using the total volume of 65.00 mL and put into an ICE table, yielding

$K_{\text{a}} = \cfrac{\left[\text{H}_3\text{O}^+\right]\left[\text{A}^-\right]}{[\text{HA}]} = \cfrac{x(0.01846... + x)}{0.01961... - x} = 1.585 \times 10^{-9}$

where $x$ is the concentration of hydronium ions at equilibrium. Solving this with the usual assumptions - that $x << 0.01846...$ - will give:

$\begin{align*} \left[\text{H}_3\text{O}^+\right] &= x = \cfrac{K_{\text{a}}(0.01961... - x)}{0.01846... + x} \\ &\approx \cfrac{1.585 \times 10^{-9} \times 0.01961...}{0.01846...} \\ &= 1.6840... \times 10^{-9}\ \text{mol L}^{-1} \\ \\ \text{Thus,} \quad \text{pH} &= -\log_{10}\left(1.6840... \times 10^{-9}\right) \\ &\approx 8.774 \end{align*}$

The same result will be obtained using the Henderson-Hasselbalch equation:

$\begin{align*} \text{pH} &= \text{p}K_{\text{a}} + \log_{10}{\cfrac{\left[\text{A}^-\right]}{\left[\text{HA}\right]}} \\ &= -\log_{10}{\left(1.585 \times 10^{-9}\right)} + \log_{10}{\cfrac{0.01846...}{0.01961...}} \\ &\approx 8.774 \end{align*}$

And there is only one small problem with this calculation and answer... it is clearly wrong! If I take a weak acid solution and neutralise less than half of it, I can't possibly get a basic solution.

So, where is the problem?

The Henderson-Hasselbalch solution is incorrect as it is derived for a system at equilibrium, and so with the concentrations of the acidic and basic forms being equilibrium concentrations. More problematically, the ICE table has started with a solution of HA and A^- and the hydronium concentration being zero. This is impossible in any aqueous system, but is usually not a problem as the contribution of the equilibrium system to the hydronium concentration is much greater than is the contribution from the auto-ionisation of the water / solvent. If you look at the calculated hydronium concentration above, you can see that it is smaller than than the amount in neutral water at room temperature. So, to find the actual pH, a quick-and-dirty solution is to add in the contribution from the solvent to the contribution from the equilibrium:

$\begin{align*} \left[\text{H}_3\text{O}^+\right]_{eqm} &\approx 10^{-7} + x \\ &= 10^{-7} + 1.6840... \times 10^{-9} \\ &= 1.016840... \times 10^{-7}\ \text{mol L}^{-1} \\ \\ \text{Thus,} \quad \text{pH} &= -\log_{10}\left(1.016840... \times 10^{-7}\right) \\ &\approx 6.993 \end{align*}$

I don't think figuring this out is a reasonable expectation at HSC level.

breh

ExtremelyBoredUser · Oct 6, 2023

SadCeliac said:
integral of cosx is sinx
integral of sinx is -cosx

fair enough tbh

yh i was just jokin, somethings never go away i.e integration skills

happily my english trauma has went away

although i doubt im as fast as i was in the hsc and i dont remember some identities

carrotsss · Oct 6, 2023

ExtremelyBoredUser said:
yh i was just jokin, somethings never go away i.e integration skills

happily my english trauma has went away

although i doubt im as fast as i was in the hsc and i dont remember some identities

how did u memorise essays without feeling extreme pain the whole time

ExtremelyBoredUser · Oct 6, 2023

carrotsss said:
how did u memorise essays without feeling extreme pain the whole time

wdym memorise i didnt lol

carrotsss · Oct 6, 2023

ExtremelyBoredUser said:
wdym memorise i didnt lol

i thought you said you did?

ExtremelyBoredUser · Oct 6, 2023

carrotsss said:
i thought you said you did?

i made essay plans and drafts but i never memorised them. id just absorb the jist of the essay and the ideas then id just practice adapting those stuff into papers.

my friends cautioned me from doing that but im built diff esp for mod C @temporarylol can affirm

can write absolute garbage but top with it

not sure if i should recommend u to memorise but if it works for u, go do it.

nsw..wollongong · Oct 6, 2023

yall i was closer with my 8.5pH than u guys were with ur 5.25 suck my d

synthesisFR · Oct 6, 2023

nsw..wollongong said:
yall i was closer with my 8.5pH than u guys were with ur 5.25 suck my d

@SadCeliac has been saying that alot too

nsw..wollongong · Oct 6, 2023

synthesisFR said:
@SadCeliac has been saying that alot too

everyone's copying me bruh

its_ace21 · Oct 6, 2023

nsw..wollongong said:
everyone's copying me bruh

they wanna be u so bad

carrotsss · Oct 6, 2023

ExtremelyBoredUser said:
not sure if i should recommend u to memorise but if it works for u, go do it.

it works well for me results-wise but I sorta hate doing it

then again I hate all english so I guess it’s not rlly much of a difference

SadCeliac · Oct 6, 2023

synthesisFR said:
@SadCeliac has been saying that alot too

i rub off on people what can I say
edit WAIT HOLD UP NO NO NO

nsw..wollongong · Oct 6, 2023

SadCeliac said:
i rub off on people what can I say
edit WAIT HOLD UP NO NO NO

stfu u copy me all the time bruh is it bc I'm adrina lima?

SadCeliac · Oct 6, 2023

nsw..wollongong said:
stfu u copy me all the time bruh is it bc I'm adrina lima?

When do I copy you huh

synthesisFR · Oct 6, 2023

SadCeliac said:
i rub off on people what can I say
edit WAIT HOLD UP NO NO NO

uve been tweakin so much recently

its_ace21 · Oct 6, 2023

SadCeliac said:
i rub off on people what can I say
edit WAIT HOLD UP NO NO NO

HSC OVERALL MARKS GOAL AND/OR REALISTIC PREDICTIONS (1 Viewer)

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