Chemistry HL/SL help

[Rigel]

Yeah well i'm doing equilibrium like on October... I'm at Bonding right now.

[master135]

Hi,

I have this problem with calculating the overall Ka value. Does anyone know how to do this? It is a triprotic acid and so it has 3 dissociation constants but I read that it is possible to obtain a general one, I just can't find how.

Thanks,

Matt

I didn't think IB covers polyprotic acids as part of equilibrium problems in its syllabus. Anyways, if I remember correctly, K overall should just be the product of the 3 different K values.

Edited June 24, 2011 by master135

[Drake Glau]

Yeah well i'm doing equilibrium like on October... I'm at Bonding right now.

It's actually covered in the acids and bases topics, not equilibrium. Ka is the disocciation constant though that gives a value to the ratio of products and reactants of the acid and it's conjugate acid. Kc...Ka...Kb, all the same just talking about a different substance is all.

Anyway, if I remember right Ka=([H⁺][A^-])/([HA])

I think that's right, I couldn't remember exactly what the CB was supposed to be, it's the other product of the disocciation because you have your hydrogen and then the anion that I can't seem to remember the name of., it's just A^-, anion, duh.

And since it's triprotic, you would simply have 3 times the H⁺ than your A^-, not that bad. Probably not in the syllabus explicitly but I'm sure it could end up on an exam

Edited June 24, 2011 by Drake Glau

[Guest kenshi64]

Hey guys, I have an overall dilemma about exam preparations and given that they're on the 14th I'd appreciate help a lot!!

So, I use the Pearson Chemistry HL, and my teachers are good, I understand what they teach in class, but sometimes, or rather frequently I'm stunned at the question that appear in the exams, I'm like, " That's too complex!!" " where did that come out of" From my previous boards I've learned that its just twisted details to a simple theory, and I was hoping I'd get some some help with "How To Get around these twisted questions and actually score in the exams" I means, I have acids and bases, which has really sinister XP XO sums in the papers. Then I have bonding and Oxidation, reduction for the first time.

Thanks, genuinely appreciate the help.

[Drake Glau]

Acids and bases you just need to calm down and assess what numbers they gave you. What equations can you use? What are they asking for? These should really be the first questions you ask yourself. Also remember that if they give you obscure values not directly related to an equation you might be able to use them to calculate the value you need. Acids and bases isn't easy, it was probably my whole classes worst topic. Between now and then I'd advise memorizing those 8 or so equations for pH pOH pKa Ka all that good stuff and remember that those can be used for bases also except then you would be using the concentrations of the BASES instead of the acids. Also remember pOH+pH=14 always and pKa+pKb=14 and Ka*Kb=10^-14 (last one was fuzzy memory, it's in your book though probably if I'm wrong).

Bonding is memorizing, unless you can think through all the e- and if you can, keep doing it.

Redox is easy, there's a few tricks for it. For the voltage potential of a cell its the E value of the reduced thing minus the E value of the oxidized thing (REDOX it's in order for you )

RED CAT, reduction...cathode...always, which means oxidation's at the anode....always.

Study the hydrogen cell. Study all your cells honestly. You can be asked draw one and it's easy points (around 4-5) if you simply remember how to draw it and label it a bit.

Anyway, back to the questions problem. Unpack it. What's given, what do they want you to answer, what calculations can you do to make your life easier, is this part B maybe there's information from part A, does part C give a hint? If you know what part C wants but not what part B remember that these questions progress as you go through them so part C will likely require information from part A and B so if you know what C wants but you're missing it completely, B is likely asking for it

[Guest kenshi64]

Acids and bases you just need to calm down and assess what numbers they gave you. What equations can you use? What are they asking for? These should really be the first questions you ask yourself. Also remember that if they give you obscure values not directly related to an equation you might be able to use them to calculate the value you need. Acids and bases isn't easy, it was probably my whole classes worst topic. Between now and then I'd advise memorizing those 8 or so equations for pH pOH pKa Ka all that good stuff and remember that those can be used for bases also except then you would be using the concentrations of the BASES instead of the acids. Also remember pOH+pH=14 always and pKa+pKb=14 and Ka*Kb=10^-14 (last one was fuzzy memory, it's in your book though probably if I'm wrong).

Bonding is memorizing, unless you can think through all the e- and if you can, keep doing it.

Redox is easy, there's a few tricks for it. For the voltage potential of a cell its the E value of the reduced thing minus the E value of the oxidized thing (REDOX it's in order for you )

RED CAT, reduction...cathode...always, which means oxidation's at the anode....always.

Study the hydrogen cell. Study all your cells honestly. You can be asked draw one and it's easy points (around 4-5) if you simply remember how to draw it and label it a bit.

Anyway, back to the questions problem. Unpack it. What's given, what do they want you to answer, what calculations can you do to make your life easier, is this part B maybe there's information from part A, does part C give a hint? If you know what part C wants but not what part B remember that these questions progress as you go through them so part C will likely require information from part A and B so if you know what C wants but you're missing it completely, B is likely asking for it

Thanks Drake! I've a few more doubts in chem, particular ones this time around haha, I'd be grateful if someone could help me with these:

Chapter: Oxidation and Reduction

'Exercises from Pearson Chem HL'

12) Draw a Voltaic cell with one half-cell consisting of Mg and a solution of Mg²⁺ ions and the other consisting of Zn and a solution of Zn²⁺ ions. Label the blah blah..

Doubt: Okay so I got the electrode and electron flow direction right, but wrote Zinc Sulfate and Magnesium Sulfate and that's wrong apparently, as its Nitrate for both! Is there a list for what anion or suffix each metal takes in such solution!? how am I supposed to know?

Doubt2 Salt bridge concept: Well, I'm told that the ions flow through this and then I'm like " won't that result in the two solutions mixing?" and then there's this whole confusion. Please tell me how this thing works.

Doubt3 Is it important to label the cotton wool tips of the salt bridge??

Have one more, Bio lecture has begun brb. Thanks!!

[Drake Glau]

Doubt#1, I got nothing. Sorry...I see no problem with this personally...

Doubt#2, Yes, they're probably mixing, but they're also separating too so there's no point in assuming you get another mixture because "theoretically" all you have are anions and no solutions at all The point of the bridge is to allow for the ions to move from one side to another. As the redox reactions occur you will eventually end up a stale cell because the side receiving the electrons will become negative and it's hard to attract electrons when you're negative The bridge allows the negative ions to go to the other side to keep the current flowing.

Doubt#3, No

[Guest kenshi64]

Here's an equation i'm suppposed to balance:

Write balanced equations for the following reactions which occur in acidic solutions:

5)MnO₄^- + H₂SO₃ ----> Mn²⁺ + SO₄^2-

Can't get the right answer, what are the steps in doing this? thanks!

Answer is : 2MnO₄^-+ 5H₂SO₃---->2Mn²⁺ + 5SO₄^2-+ 3H₂O +4H⁺

[Drake Glau]

MnO₄^- + CH₃OH --> Mn²⁺ + CO₂ is where you start.

Half reactions are like taking it apart, the parts that look like they go together. 2 halves make a whole

MnO₄^- --> Mn²⁺

First thing is balancing the O by adding water from the surrounding aqueous solution. (actually the FIRST thing is balancing anything but O and H by doing the normal equation balancing with coefficients, O and H are special )

MnO₄^- --> Mn²⁺ + 4H₂O

Now you need to balance the hydrogen from the water by adding H⁺

8H⁺ + MnO₄^- --> Mn²⁺ + 4H₂O

Now your charges don't balance, so we need to do that. By adding some electrons. You have +7 on the left and +2 on the right, so add 5 electrons to the left.

5e^- + 8H⁺ + MnO₄^- --> Mn²⁺ + 4H₂O

Ok, that one is done now.

Your other equation would be the methanol and CO₂

CH₃OH --> CO₂

Your oxygen and hydrogens are now unbalanced, and you balance oxygen with waters (they come from the aqueous solution that the reaction is happening in)

H₂O + CH₃OH --> CO₂

Now for hydrogens, 6 on the left, none on the right, so the right side gets 6 H ions.

H₂O + CH₃OH --> CO₂ + 6H⁺

Charges are now off, 0 on the left and +6 on the right so the right side gets 6 electrons

H₂O + CH₃OH --> CO₂ + 6H⁺ + 6e^-

Now we have both half reactions done, we get to combine them. When you combine the reactions the electrons MUST cancel out (this is much like how you do hess's law by the way).

H₂O + CH₃OH --> CO₂ + 6H⁺ + 6e^-

5e^- + 8H⁺ + MnO₄^- --> Mn²⁺ + 4H₂O

One has 6 electrons, one has 5 electrons, least common multiple of 30 (sadly ) so your methanol reaction needs to be multiplied by 5 and your permanganate equation gets to be multiplied by 6

5H₂O + 5CH₃OH --> 5CO₂ + 30H⁺ + 30e^-

30e^- + 4818H⁺ + 6MnO₄^- --> 6Mn²⁺ + 2419H₂O

Now we start canceling stuff out just like hess's law. The electrons are 30 and 30, they're gone. Next we'll do water, 5 and 24, 24-5=19 waters on the right hand side. 48 hydrogen ions on the left, 30 on the right, so 18 are left on the left side.

So what we have left is. 6MnO₄^- + 5CH₃OH + 18H⁺ --> 5CO₂ + 6Mn²⁺ + 19H₂O

And wow the amount of bbc code in this is amazing, I'm hoping this works...

You can make steps for doing this if you have troubles on where to start, it's like a 7 step process for both reactions, just do a couple more and you'll quickly pick up on it

This post took too long to make the first time, it's not your problem but it outlines everything you need to do with an example

TOO much bbc code to do it again

[-Carl]

Would someone mind explaining what Solution Equilibrium is? Can't seem to grasp it.

[Drake Glau]

It's just the state of equilibrium for a solution. The concentrations of the reactant and product are not changing because the forward reaction and backward reaction are at the same rate so the forward reaction is creating products just as fast as the backward reaction is using them.

[-Carl]

Perhaps my question was a bit vague. It states the rate of dissolving the solute crystals equals the rate of crystallizing of the dissolved solute particles, what exactly is the rate of "crystallizing"? Also, regarding the equilibrium constant "K", it states a large value of K means that most of the reactants were converted into products before equilibrium was reached. I don't understand the whole "before equilibrium was reached"? Isn't this value being calculated AT equilibrium?

[Drake Glau]

Oh, right, solution. Sorry

By crystallization they are talking about the dissolved "solid" returning to it's solid state. Take some sugar and dump it in some water. After a while you get a supersaturated solution and there are sugar crystals on the bottom of the cup. This isn't necessarily because the water is holding everything it can but rather that rate at which the sugar is recrystallizing on the bottom of the cup is equal to the rate that the sugar already on the bottom of the cup is dissolving.

It is, equilibrium simply means that the forward and reverse reactions are occurring at the same rate. The K value is the ratio of concentrations however so at equilibrium the concentrations can be near equal, really high, or really low even. Since K is [products]/[reactants] a K value much greater than 1 would imply that the concentration of the products is extremely high relative to the reactants so it is assumed that the reaction just goes to completion. If it is close to one, but still greater than one, then there is a higher concentration of products than there is reacants so equilibrium is positioned slightly to the right (forward reaction). If K is between about .2 and 1 then the equilibrium position is to the left because the ratio implies that the concentration of the product is less than the concentration of the reactants. If K=0 there is no reaction going on because if K=0 then there are no products being formed at all.

Remember that equilibrium is equal rates, not equal concentrations.

To address the "before equilibrium was reached", it is simply saying that the reaction made more products to achieve the K<1 before the rates were equal.

[dessskris]

It states the rate of dissolving the solute crystals equals the rate of crystallizing of the dissolved solute particles

the rate of dissolving the solute crystals = the rate of, say, forward reaction

the rate of crystallizing of the dissolved solute particles = the rate of backward reaction

you see when crystals dissolve in a solution then the opposite would be the dissolved particles crystallising back.

[-Carl]

Thank you to both Glaus.

I'm having trouble identifying amphoteric substances. For example in the multiple choice question "Which of the following is likely to be an amphoteric substance?" a)HCl b) NaOH c)NH3 d)Na2HPO4 . I'm not quite sure which one it is, I know that amphoteric substances can act as both acids and bases, but how do you identify which one it is based on just being given their formulas? Also, if someone could kindly explain why the other answers are incorrect that would be great. I'm not sure if this is covered in the IB syllabus, but I'm doing this for preparation for IB2, so I'm assuming it is

Edited August 2, 2011 by -Carl

[Keel]

Thank you to both Glaus.

I'm having trouble identifying amphoteric substances. For example in the multiple choice question "Which of the following is likely to be an amphoteric substance?" a)HCl b) NaOH c)NH3 d)Na2HPO4 . I'm not quite sure which one it is, I know that amphoteric substances can act as both acids and bases, but how do you identify which one it is based on just being given their formulas? Also, if someone could kindly explain why the other answers are incorrect that would be great. I'm not sure if this is covered in the IB syllabus, but I'm doing this for preparation for IB2, so I'm assuming it is

Hi,

I would go for D. When doing multiple choice questions you can usually eliminate half of the answers as unacceptable. I would first eliminate A and B. This is because they are either a strong base or acid, it is ulikely that NaOH will act as an acid thus you are left with C and D. NH₃ is a weak base, for it to act as an acid it would need to donate a proton making the resulting molecule unstable. However, if you look at Na₂HPO₄ it has Na and H atoms. If it were to be and acid, it could simply donate a proton resulting in [Na₂PO₄]^- which is relatively stable. If it were to act as a base and accept protons it is likely that a Na will be replaced by a H giving NaH₂PO₄ or H₃PO₄. Thus I would choose D. I guess a simplier way of thinking about it is, can it act as an acid i.e. Does it have H atoms? Can it act as a base i.e. Can it accept H atoms? Then choose the best option.

[Summer Glau]

Hey, how would you calculate the pH of a substance given its molarity? Say I have a 250mL, 0.5 M solution of HCl. How do I find out what the pH is?

Thanks

[dessskris]

pH = -log₁₀[H⁺]

if I remember correctly.

and [H⁺]=[HCl]

Edited August 29, 2011 by Desy Glau

[Emmi]

Hey, how would you calculate the pH of a substance given its molarity? Say I have a 250mL, 0.5 M solution of HCl. How do I find out what the pH is?

Thanks

I love acids and bases

HCl is a strong acid. Because of that, the molarity is equal to the hydrogen ion concentration.

M = [H₃O⁺] = 0.5

Now take the negative log of the hydrogen ion concentration (0.5) which will give you the pH of your HCl solution

pH = -log(0.5) = 0.301 (rounded to three decimal places)

[Guest KAPOWW!!]

Yo guys, what's the best way to prepare for Chemistry Equilibria and Enthalphy- Those are probably not the chapter names, but you get my flow

Since my grades dropped because of this chapter as I didn't know what to concentrate on. Much appreciated.