Chemistry Question Thread

God Incorrect. pKa is not a measure of pressure. They don't teach it in VCE but it's -log10 of an acidity constant.

purpleunicorn
pKa is actually the pH at which 50% of an acid is deprotonated eg since ethanoic acid's pKa is 4.74, at a pH of 4.74 you'd expect half of the species present to be unionised ethanoic acid (CH3COOH) and half to exist in the form of ethanoate ions (CH3COO-). It's defined as -log10(Ka), where Ka is the acidity constant i.e. equilibrium constant of an acid dissociation reaction

I’m so sorry… kPa ≠ pKa .

Thanks Billzene

purpleunicorn correct, the pKa is the halfway point between equivalence points, you see them as flat parts on titration curves since when pH = pKa, the buffer system is the most resistant to pH changes according to the Henderson-Hasselbach equation taught in uni chem (not sure if they teach it in QCE)

Sure thing!

So ur electrolytic cell uses a dc current to force a non spontaneous redox reaction.

The cathode is the sight of reduction. It’s mass will increase, as the electrons are donated to the ions, plating it in metal. Hence ‘electro-plating’

Electrons flow in the opposite direction to conventional current. They flow to the cathode. Hence the one on the right would be the cathode.

There exists a E0 for each redox reaction, which (for electrolytic cells) is the voltage needed for the reaction to occur (actually it’s anything greater than E0)

However, if the voltage gets too high, then side reactions might occur. These aren’t good (as they can make it impure/release gases and stuff)

Part D seems like a hard-to-answer question. Current is conducted through the wire I guess, and through the movement of ions in the solution? I’m not sure what they are after there…

Hope this helps!

jinx_58 for part d, maybe they're asking for the conventional current through the wire which is opposite in direction to e- flow. e-s go from anode to cathode, so you'd draw conventional current from cathode to anode since that's supposed to represent flux of positive charge. The electrolyte is part of the circuit as well, since cations are attracted to the cathode, you'd draw conventional current in the electrolyte going from anode to cathode

Judging from your opinions of the chem exam this year. What would you say is the range of marks out of 120 for 47ss given rank 1 in sacs.

Hi there,

I just wanted to ask how to answer some chem questions, but one of them involves an image do I don't know how to add the link because when I don't spits out pages of code? Does anyone know how I could share the images without this issue?

Bibliii maybe use google drive and turn sharing to everyone with link can view?

Billzene ahhh my google drive has been playing for a few days, but I'll just put up the ones that I can type:

16. Calculate the pH of a 6.50 x 10^-3M KOH solution? (Also for this what does KOH mean??)

17. Calculate the pH of a 0.227M solution of hydrocyanic acid (HCN). The Ka=6.2 x 10^-10 (Alsp is Ka the same as pKa.)

18. 31.8g of Na2CO3 was dissolved to make a 500mL solution. This was diluted by pippetting 5mL of this solution into a 50mL volumetric flask and diluting to volume with deionised water. This 50mL was titrated and the results showed that the solution completely reacted with an unknown concentration of 16.8mL of H2SO4. What was the concentration of H2SO4 to 2 decimal places?

Could anyone explain these to me in steps, thank youuu. Also I know these are all extremely simple questions, but my head has been having a hard time with readjusting to school and having anxiety (funnnn). So yeah, sorry about that everyone.

Bibliii KOH as God notes is the strong base potassium hydroxide, potassium is denoted by the symbol K since it’s called kalium in Latin. To approach this question, I’d take -log10( ) of the concentration of KOH they gave you to get the pOH, and then rearrange the formula pOH + pH = pKw = 14 for pH

For 17, hydrocyanic acid is a very weak acid judging from its low Ka magnitude so set up an ICE table as follows

Initial [HCN] = 0.227 M, initial [CN-] = 0 M, initial [H+] = 0 M
Change [HCN] = -x M, change [CN-] = +x M, change [H+] = +x M
Equilibrium [HCN] = 0.227 - x M, equilibrium [CN-] = x M, equilibrium [H+] = x M

Then equate the magnitude of Ka with your Ka expression

6.2 x 10^-10 = (x M)² / (0.227 M)*

*In uni chem here in Victoria we’re always allowed to assume that the concentration change of an weak acid is 0, if QCE doesn’t allow it, you’ll have to use the quadratic formula.

Now solving for x should give you [H+], which you should then convert into pH

Ka isn’t the same as pKa, pKa is -log10(Ka) (whenever you see p in chem notation, it means -log10( ), such that pH = -log10(H+) and pOH = -log10(OH-))

18 is a titration question that involves a dilution, always start with a balanced equation and remember that your standard was diluted by a factor of 10.

So you know the original concentration of your standard was 31.8 g / 106 g/mol = 0.3 mol in 500 mL = 0.6 M. Diluting that by a factor 10 (use c1v1 = c2v2, where c1 = 0.6 M, v1 = 5 mL, c2 is your unknown and v2 = 50 mL). Your DILUTED concentration would be 0.06 M

Now you know 50 mL of a 0.06 M Na2CO3 solution reacted with 16.8 mL of H2SO4 solution. The balanced equation is pretty simple since everything is 1:1:1:1:1 (Na2CO3 + H2SO4 —> Na2SO4 + H2O + CO2). Find n(Na2CO3), which will be equivalent to n(H2SO4) according to the balanced molar ratio. Then use c = n/v to find [H2SO4], remember that v must be in L so 16.8 mL = 0.0168 L.

Sorry guys, after 3 months off I’ve suddenly become completely inept at Chem

Thanks @Billzene and @PhytoPlankton

Much much better responses