A Chemistry thread of the QCE students, feel free to ask any questions!

Anyone is welcome to answer 🙂

3 months later

Edit: this is wrong

pKa is pressure. (kilopascals) Completey unrelated to pH - which is a measure of the acidity of a solution.

    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

          Billzene

          Is that why in a titration graph, the half equivalence point = pKa = pH?

          idk
          i get it but don't get

              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)

                Yes @Billzene is correct. Specifically in QCE, all you need to know is pKa in terms of titration graph (you need to know why as QCE can ask for an explanation) and the Henderson-Hasselbalch equation like Billzene said above. You'll also need to know how to calculate the Ka by using Ka=[H3O+][A-]/[HA] and then the pKa from it which is a simple -log.

                Hope that helped.

                • PP
                21 days later

                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!

                  Thank you God

                  Appreciate your help mate!

                  -jinx_58

                  • God likes this.

                    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

                        8 days later

                        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.

                        2 months later

                        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?

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

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

                          2. 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.)

                          3. 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

                              Hi 🙂

                              1. KOH is potassium hydroxide, which is a base. (When in solution, it goes to OH- and K+ ions)

                              So [OH-] = [KOH]

                              And then [H+] = 10^-14/[OH-]

                              And then ph= -log 10 ([h+]}

                              (I could be slightly wrong there, I am a little rusty)

                              I can’t do 17 since I did VCE (vic high school) and we didn’t have Ka. Billzene might be able to help though)

                              And Q18 is giving me serious ptsd just looking at it , and i might have a crack tomorrow. 😉

                              Night!
                              -G

                                Bibliii
                                Hey Bibliii,

                                I think QCE does it a different way so here are the solutions to all three questions in "QCE" terms (correct me if I'm wrong):
                                16. KOH -> K+ + OH-
                                So, as @God stated, there is a 1:1 molar ratio so [OH-] = [KOH] = 6.50 x 10^-3 M
                                Since pOH = -log[OH-] = -log(6.50 x 10^-3 M) = 2.19
                                So, to find pH = 14 - pOH = 14 - 2.19 = 11.81

                                17. HCN -> H+ + CN-
                                Ka = [H+][CN-]/[HCN] = 6.2 x 10^-10 (pKa is just -log(Ka))
                                [H+][CN-]/[HCN] = (x)(x)/(0.227 - x) = 6.2 x 10^-10 and we usually assume x to be quite small so we take it as negligible (or we can calculate it with it but there's no point of wasting time for it).
                                Thus, x2/0.227 M = 6.2 x 10^-10
                                x2 = 1.407*10^-10
                                x = 1.19 x 10^-5
                                [H+] = 1.19 x 10^-5
                                pH = -log[H+] = -log(1.19 x 10^-5) = 4.93

                                18. The molar mass of Na2CO3 = 105.99 (you can calculate this through the periodic table but I searched it up from Google because I was lazy).
                                So since the number of moles (n) = mass/molar mass, the number of moles (n) of Na2CO3 = 31.8 g/105.99 g mol^-1 = 0.30 mol
                                This means concentration (c) = n/V = 0.30 mol/0.5 L = 0.60 M (concentration of Na2CO3 in 500 mL solution).
                                The dilution factor is 10 because it is from 5 mL solution into 50 mL (this can also be worked out from C1V1=C2V2) so this means the concentration of Na2CO3 in 50 mL solution = 0.06 M.
                                For the titration part, your best friend is C1V1=C2V2 so all you have to do is sub in all the values. The LHS will be the base part (Na2CO3) and the RHS is the acid part (H2SO4).
                                C1V1=C2V2
                                0.06 M x 50 mL = C2 x 16.8 mL (you should convert all mL->L like in the concentration calculation at the start but I ignored it here because it would be the same answer).
                                C2 = 0.06 x 50 / 16.8 = 0.18 M
                                Therefore, the concentration of H2SO4 is 0.18 M.

                                Make sure to always write the reaction equation at the start of the question - that's very important. This is how I was taught to calculate when I did QCE but anyone, let me know if anything's wrong because it's been a while. Hope that helps!

                                • PP