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.

    PhytoPlankton lol got beaten by you, I can confirm that you ain’t rusty and everything is correct here even though I personally would turn 50 mL into L since the concentration (0.06 mol/L) can only cancel out litres in dimensional analysis when following the c = n/v formula, and L is the most used volume unit in chem

      Ahh thank you very much everyone for the help!!

      6 days later

      Hey everyone!

      Balancing equations are kicking my ass rn and there are a few that I'm stuck on. Any help really would be greatly appreciated!

      How would you balance this equation?
      ___ HCI + ____ Ca(OH)2 = ______ CaCi2 + H2O

      Thank you in advance! 🙂

        sodacat_
        Hey sodacat,

        I believe you meant CaCl in the products' side of your equation so in that case, below is the balanced equation:
        2HCI + Ca(OH)₂ = CaCl₂ + 2H₂O
        Also, make sure you don't superscript the 2's and subscript it where you can (but that's quite insignificant 😃).

        Hope this helps and feel free to ask more questions!

        • PP

            sodacat_
            Just adding to PhytoPlankton's excellent response with an example methodology you can use when answering these types of questions.

            What I find really helpful when balancing equations is to first balance elements other than hydrogen and oxygen (in this case calcium and chloride), then balance the hydrogens and finally balance the oxygens.

            1. Balancing calcium and chloride
              We have equal numbers of calcium atoms on both sides of the equation so we don't have to do anything.
              There are two chloride atoms on the right hand side of the equation and one chloride atom on the left hand side. So we need to double the number of chloride atoms on the left hand side by putting a '2' in front of HCl.

            2HCI + ____ Ca(OH)2 --> ______ CaCl2 + H2O

            1. Balancing hydrogen
              We have four hydrogen atoms on the left hand side (2 from HCl and 2 from Ca(OH)2) but only two hydrogen atoms on the right hand side of the equation. Therefore we need to double the number of hydrogen atoms on the right hand side. We do this by putting a '2' in front of H2O.

            2HCI + ____ Ca(OH)2 --> ______ CaCl2 + 2H2O

            1. Balancing oxygen
              We have two oxygen atoms on the left hand side (from Ca(OH)2) and two oxygen atoms on the right hand side. Oxygen atoms are balanced so we don't need to do anything.

            2HCI + Ca(OH)2 --> CaCl2 + 2H2O

            Hope that helps.

                PhytoPlankton

                GreenAcorn

                Thank you both! Unfortunately, I did not see these responses until after I had already figured it out and got a little help from my chem teacher (flat laptops are fatal lol), but the responses were very effective in helping me to understand it all!

                    13 days later

                    Hiii! Data test in two days. I'm not really sure of when to use pka, pkb, kb, ka, and h30+ and oh- ions, sooo is anyone able to give a lil brief on what questions to use these in?

                      Bibliii
                      pKa -> Kinda like pH, except for a weak acid
                      pKb -> Kinda like pH, except for a weak base
                      Know the difference between a weak and a strong acid, if you don't know this, feel free to ask.
                      If you're given the concentration of Ka or Kb, you can use it to find pKa or pKb, i.e. the strength of the acid or base
                      You might be given an equilibrium equation, and you'll have to figure out which one is the weak acid or base. Memorise the obvious ones like ethanoic acid (weak acid), hydrochloric acid (stronk acid), sodium hydroxide (stronk base), Ammonia (weak base)

                      Kb -> equilibrium constant of weak base
                      Ka -> equilibrium constant of weak acid

                      H3O^+ ions -> pH = [H3O^+], same as H^+, hydronium ions in acid
                      OH^- ions -> pOH = [OH^-], hydroxide ions in a base
                      In an acidic or basic solution, using these concentrations will help to find the pH or pOH, that is the power of hydrogen or hydroxide.

                      pOH + pH = 14 -> Handy trick

                      Any questions, feel free to ask 🙂
                      -jinx_58

                          Bibliii

                          Ka = the equilibirum constant of an acid dissociation reaction eg for the equilibrium system HCOOH + H2O <-> HCOO- + H3O+, Ka = [H3O+][HCOO-] / [HCOOH] (remember you exclude water for aqueous equilibria since its concentration change is negligible)

                          pKa = pH above which half of all acid molecules are deprotonated. Given by -log10(Ka), so higher pKa = weaker acid. If you inspect a titration curve, they're the half way points between equivalence points. They're really flat since when pH = pKa, an acid buffer system is most resistant to change according to the Henderson-Hasselbach equation (I don't know to what extent banana benders need to know this, I only learned Ka in uni chem in Victoria). You can memorise the property of pKa by the fact that when pH is above pKa, the equilibrium system is deficient in H+. According to Le Chat's principle, the system will want to partially oppose that, which it will by having your acid donate its proton.

                          Kb = the equilibrium constant of a base ionisation reaction eg the Kb for NH3 + H2O <--> NH4+ + OH-, Kb = [OH-][NH4+] / [NH3]. Kb and Ka can be generalised as [ionised species] / [unionised species]

                          pKb = the pH above which a base will be protonated. Similar story with pKa, when pH is above pKb, the system knows there's not enough H+. To compensate, you'll want less OH- in solution, forcing the base ionisation reaction to proceed in reverse

                          H3O+ = the protonated form of water, ie its conjugate acid. Abundant in acidic conditions since the acid has donated its H+ to the solvent molecule (H2O).

                          OH- = the deprotonated form of water ie its conjugate base. Abundant in basic conditions since the base has stolen H+s from H2O.

                          One small correction to jinx's excellent response, technically strong acids also have Ka and pKa values despite the fact that strong acid ionisation is pretty irreversible. For example, we are required to know in uni that pKa of HCl is -5 for purposes of comparing leaving group strength in organic reactions. Ditto for strong bases with regards to Kb and pKb

                          I don't know how QCE structures their data tests so someone else can help with that

                            Bibliii

                            Yes, they may ask you! Have you done any practice questions for this topic, if so it's likely to be in that format!

                            Also, you should make sure you do questions where you predict the colour of the solution, this is something I kinda struggled with it in the exam because I didn't prepare as well!

                            Hope this helps! Any questions feel free to ask as I completed QCE last year!

                              7 days later

                              Hi guys,

                              I need some help. So I'm supposed to work out an equation for Sodium thiosulfate reacting with Hydrochloric acid. My teacher told me to search it up and so I did. When I did, it came up with Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + SO2(g) + S(s). Then, in another question, it said to write a balanced equation but THAT IS THE BALANCED EQUATION!!! I think. So what do I do? Do I just write down the balanced equation only?

                              Thanks so much, Leeshi! 😄

                              Nvm, I figured it out 😅 but just want to check. Is the equation for Sodium thiosulfate reacting with Hydrochloric acid Na2S2O3(aq) + HCl --> NaCl+H2O+SO2+S

                              Please don't mind the numbers, they're supposed to be subscript ;-;

                              Thanks, Leeshi.

                                7 days later

                                Heyoo,

                                I have my student experiment and my group decided to see what happens to conductivity when the temperature of ethanoic avid solution is increased. I've just got a tiny section of sample calculations to add before I move in my report. Would calculating Ka knowing the concentration H3O+ ions look like this (I'll try and type it out as best I can):
                                Ka = [H3O+][CH3COO-]/[CH3COOH]
                                Apparently CH3COO- concentration is equal to H3O+ so I think that explains the next step...
                                = [H3O+][H3O+]/[CH3COOH]
                                = [H3O+]2/[CH3COOH]

                                I have the concentration of CH3COO and also the concentration of H3O+ (via our experimental pH values) so substituting that in should be all good to go?

                                  chocomilkenthusiast Good thinking with regards to the formulas, you can also use an ICE table if you’re unsure. I’d like to draw your attention to the fact that Ka is an equilibrium constant, which means it’s dependent on the endo/exo status of the ionisation reaction, so you may need some literature enthalpy values (theoretical accurate values) to assess your accuracy. To complicate things further, the enthalpy value changes depending on the ambient temperature. Just some things to think about here

                                    lucy__ FYI the changing enthalpy based on changing temperature thing requires integral calculus (Kirchoff’s Law), so for the sake of your sanity, just look up some values of heat of ionisation for ethanoic acid at different temperatures.

                                    The reason why I had to talk about it is because normally temperature changes don’t incur a sign change in the enthalpy change as in high school you assume heat capacities to be constant regardless of temperature. However, heats of ionisation (enthalpy change for an acid dissociation reaction) are so small that even the smallest temperature change may result in a new deltaH with an opposite sign.

                                    Also, most literature refers to ethanoic acid as “acetic acid”, you’d be surprised how you can’t find anything with the proper IUPAC name but find a million things with its common name.