Chemistry Question Thread

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

Ahh thank you very much everyone for the help!!

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_
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

2. 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

3. 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!

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

jinx_58

Thanks! Could they ask about conjugate acid base pairs in a data test?

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.

Leeshi you have 2Na on LHS but only 1Na on RHS, so the first equation you gave was correct

Billzene
Thank you for the feedback!

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?