HSC 2016 Chemistry Marathon (1 Viewer)

mrstripedshades · May 11, 2016

Re: HSC Chemistry Marathon 2016

DatAtarLyfe said:
Edit: i was meant to say the answer is D, 0.38kg

Whats the equation you use to get the enthalpy? Just wondering if there is one to remember it not really familiar with hose kind of questions.

Occupied · May 13, 2016

Re: HSC Chemistry Marathon 2016

mrstripedshades said:
Whats the equation you use to get the enthalpy? Just wondering if there is one to remember it not really familiar with hose kind of questions.

Triangle H = -MC triangle T represents:
Change in heat = - Mass * Heat capacity of water * Change in temp

Flop21 · May 14, 2016

Re: HSC Chemistry Marathon 2016

How do I calculate the number of moles of H+ neutralised??

I have the number of moles of HCl in a solution (calculated from the molarity * volume), same with NaOH. The limiting reagent is HCl.

Now is the number of moles of H+ neutralised the moles of the limiting reagent or?

leehuan · May 14, 2016

Re: HSC Chemistry Marathon 2016

Flop21 said:
How do I calculate the number of moles of H+ neutralised??

I have the number of moles of HCl in a solution (calculated from the molarity * volume), same with NaOH. The limiting reagent is HCl.

Now is the number of moles of H+ neutralised the moles of the limiting reagent or?

In each mol of HCl there is exactly one H⁺

There is no reason for the moles of hydrogen ions to unnecessarily be different to the moles of hydrochloric acid present.

packpig123 · May 14, 2016

Re: HSC Chemistry Marathon 2016

Calculate the pH of a 0.21 M solution of acetic acid (CH3COOH).
pKa=4.76
answer=2.72
I know for a base its 14-pKa and then (M.10^-(14-pKa)^1/2 etc but whats the formula for the one above

leehuan · May 14, 2016

Re: HSC Chemistry Marathon 2016

packpig123 said:
Calculate the pH of a 0.21 M solution of acetic acid (CH3COOH).
pKa=4.76
answer=2.72
I know for a base its 14-pKa and then (M.10^-(14-pKa)^1/2 etc but whats the formula for the one above

In my entire chemistry life I have not seen pKa mentioned once. (Which means I highly doubt it's required for the HSC course.)
What even is it.

InteGrand · May 14, 2016

Re: HSC Chemistry Marathon 2016

leehuan said:
In my entire chemistry life I have not seen pKa mentioned once. (Which means I highly doubt it's required for the HSC course.)
What even is it.

K_a is the acid dissociation constant for an acid. pK_a is the colog (negative log-base-10) of that (by definition of "p"). K_a is a quantitative measure of the strength of an acid in solution. It's not in the HSC Chemistry syllabus (unfortunately!) though – it would've helped explain acid chemistry a lot better had it been in it. The smaller the K_a value (equivalently, the larger the pK_a value), the weaker the acid.

E.g. here are some K_a values for some common acids:

Hydrofluoric acid - KHF: 3.5 x 10^-4
Sulfuric acid - KH₂SO₄: 1.2 x 10^-2
Acetic acid - KCH₃COOH: 1.8 x 10^-5
Phosphoric acid- KH₃PO₄: 7.5 x 10^-3.

This tells us for example that out of these, acetic acid is the weakest, and sulfuric acid is the strongest acid.

Further reading: https://en.wikipedia.org/wiki/Acid_dissociation_constant .

Colog: https://en.wikipedia.org/wiki/Cologarithm .

InteGrand · May 14, 2016

Re: HSC Chemistry Marathon 2016

packpig123 said:
Calculate the pH of a 0.21 M solution of acetic acid (CH3COOH).
pKa=4.76
answer=2.72
I know for a base its 14-pKa and then (M.10^-(14-pKa)^1/2 etc but whats the formula for the one above

$\noindent Let the $\text{K}_{\text{a}}$ be $K$. We know that $K = 10^{-\text{pK}_{\text{a}}} = 10^{-4.76}$.$

$\noindent Writing out the formula for $K$, we have $K=\frac{\left[\text{A}^{-}\right]\left[\text{H}^{+}\right]}{\left[\text{HA}\right]}$. Now, both the $\left[\text{H}^{+}\right]$ and $\left[\text{A}^{-}\right]$ values for an acid already in solution will be zero at the time of dissociation and $x$ say once dissociation is complete. Also, $\left[\text{HA}\right]$ will decrease from $0.21$ to $0.21 -x$ over this time frame. So putting these into the equation for the constant $K$, we have$

$\begin{align*}K&= \frac{x\cdot x}{0.21-x}.\end{align*}$

$\noindent Rearranging this yields the quadratic equation $x^2 + Kx -0.21 K = 0$. Using the quadratic formula and taking the positive root, we have $x=\frac{-K+\sqrt{K^2 + 4 \times 0.21K}}{2}=0.00190166626\ldots \equiv X$, say. Now, the pH is just the negative log-base-10 of this, i.e. $\text{pH}=-\log _{10} X = 2.720865\ldots \approx 2.72$.$

This method is explained here: http://classroom.synonym.com/determine-ph-pka-2832.html .

packpig123 · May 15, 2016

Re: HSC Chemistry Marathon 2016

InteGrand said:
$\noindent Let the $\text{K}_{\text{a}}$ be $K$. We know that $K = 10^{-\text{pK}_{\text{a}}} = 10^{-4.76}$.$

$\noindent Writing out the formula for $K$, we have $K=\frac{\left[\text{A}^{-}\right]\left[\text{H}^{+}\right]}{\left[\text{HA}\right]}$. Now, both the $\left[\text{H}^{+}\right]$ and $\left[\text{A}^{-}\right]$ values for an acid already in solution will be zero at the time of dissociation and $x$ say once dissociation is complete. Also, $\left[\text{HA}\right]$ will decrease from $0.21$ to $0.21 -x$ over this time frame. So putting these into the equation for the constant $K$, we have$

$\begin{align*}K&= \frac{x\cdot x}{0.21-x}.\end{align*}$

$\noindent Rearranging this yields the quadratic equation $x^2 + Kx -0.21 K = 0$. Using the quadratic formula and taking the positive root, we have $x=\frac{-K+\sqrt{K^2 + 4 \times 0.21K}}{2}=0.00190166626\ldots \equiv X$, say. Now, the pH is just the negative log-base-10 of this, i.e. $\text{pH}=-\log _{10} X = 2.720865\ldots \approx 2.72$.$

This method is explained here: http://classroom.synonym.com/determine-ph-pka-2832.html .

Thanks!

Flop21 · May 29, 2016

Re: HSC Chemistry Marathon 2016

Can someone double check my answers and help me with the one unanswered? They're the functional groups.

leehuan · May 29, 2016

Re: HSC Chemistry Marathon 2016

Flop21 said:
Can someone double check my answers and help me with the one unanswered? They're the functional groups.

The unanswered one looks like butanoic acid so

(But the HSC course doesn't give proper references to ketones and aldehydes)

Shuuya · May 29, 2016

Re: HSC Chemistry Marathon 2016

Flop21 said:
Can someone double check my answers and help me with the one unanswered? They're the functional groups.

The unanswered one is a carboxylic acid.

Flop21 · May 29, 2016

Re: HSC Chemistry Marathon 2016

Thanks are the rest correct?

leehuan · May 29, 2016

Re: HSC Chemistry Marathon 2016

Flop21 said:
Thanks are the rest correct?

leehuan said:
The unanswered one looks like butanoic acid so

(But the HSC course doesn't give proper references to ketones and aldehydes)

This implies I don't know

(wait for a proper scientist not me to show up)

BlueGas · May 29, 2016

Re: HSC Chemistry Marathon 2016

leehuan said:
The unanswered one looks like butanoic acid so

(But the HSC course doesn't give proper references to ketones and aldehydes)

Ketones and aldehydes are mentioned in biology (uni level), and her questions wasn't HSC targeted as it's a university question

leehuan · May 29, 2016

Re: HSC Chemistry Marathon 2016

BlueGas said:
Ketones and aldehydes are mentioned in biology (uni level), and her questions wasn't HSC targeted as it's a university question

You'd think I have figured that given Flop is in our cohort.

It just means posting it here (in the HSC marathon) might not guarantee an answer.

InteGrand · May 29, 2016

Re: HSC Chemistry Marathon 2016

Uni-level Q's should probably be posted in the university Science threads in order to maximise chances of getting the question answered.

Flop21 · May 29, 2016

Re: HSC Chemistry Marathon 2016

I'm doing intro chem which is equivalent HSC chem, so hence why I've been posting here.

Cool seems we're at uni level now.

someth1ng · May 29, 2016

Re: HSC Chemistry Marathon 2016

Flop21 said:
Can someone double check my answers and help me with the one unanswered? They're the functional groups.

Top middle is a ketone. The -COOH denotes a carboxylic acid.

wesadora · Nov 2, 2016

Re: HSC Chemistry Marathon 2016

5.00g table salt was dissolved in a 250L flask. He took a 10mL sample and diluted it to 1L for analysis. The absorbance was measured at 0.804.

From graph, 0.804 corresponds to 50ppm of sodium (50mg/L). Plz halp

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