General Chemistry Question Thread 2

Ludacris said:

Can you please explain the effective nuclear charge a bit? Thanks

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Basically, the nucleus is really positively charged, and attracts electrons which have a negative charge. However, the valence electrons (the outermost electrons) might not feel the full "pull" of the electrons, since the inner electrons are blocking the pull. It's like if you had a magnet and then surrounded it with paper... the magnet wouldn't pull stuff as hard since the paper is "blocking" the magnetic field. In this case, the inner electrons are blocking the "pull" that the outer electrons feel.

Higher nuclear charge lowers orbital energy, because it is a stabilizing force (it increases the electron-nucleus attraction). What this will do is make the electrons more difficult to remove, since the outermost electrons that would be removed first are being firmly pulled by the nucleus. If there's a lot of shielding, those outer electrons don't feel the pull and can be removed more easily.

If you have specific questions related to effective nuclear charge, let me know! Hope this helps.

Can you please explain how an indicator works in titrations? What are they and what do they do?
For the MCAT, what's important to know about them.. they're represented by the symbol HIn right?

Thanks

axp107 said:

Can you please explain how an indicator works in titrations? What are they and what do they do?
For the MCAT, what's important to know about them.. they're represented by the symbol HIn right?

Thanks

Click to expand...

Yes, they are. The HIn is the protonated indicator. It has an H+ on it. The In- is the deprotonated version. Basically, that proton has a certain acidity or basicity associated with it... just so happens that these indicators are unique in that when they go from HIn to In-, there is a color change.

Let's assume you have a sample of acid, and you're adding base. First the sample of acid gets deprotonated by the base, and then when the acid runs out, the indicator deprotonates, and changes colors. This makes it super easy to know that your reaction is done, because BAM! color change. If there were no color change, you wouldn't know if you ran out of acid or not (since the acid is typically clear, and the base is clear). The indicator just serves as a little notification that the titration is over.

What's important is you need the right indicator for the job. You can't use an indicator that doesn't match with your reaction.

Indicators don't change color sharply at one exact pH, but rather change over a small range of pH. When do you use a specific indicator? This website has some graphics (scroll down) that show you how indicators are matched. You want the pKa of the indicator to be the pH of the equivalence point.

Thanks for the help. Since this also had to do with acid/base titrations, I figured I'd ask... normally when you titrate a monoprotic acid with a base, at the half equivalence point, pH = pka. What if you titrate a base with an acid, would the half equivalence point be when pH= pkb instead? Thanks

axp107 said:

Thanks for the help. Since this also had to do with acid/base titrations, I figured I'd ask... normally when you titrate a monoprotic acid with a base, at the half equivalence point, pH = pka. What if you titrate a base with an acid, would the half equivalence point be when pH= pkb instead? Thanks

Click to expand...

The half-equivalence point you're talking about comes from the Henderson-Hasslebalch equation:

pH = pKa + log([base]/[acid])

So you said half equivalence point, which is when the [HA] equals to the [A-], and of course vice-versa.

When that happens, [HA]/[A-] has to be 1, since they are equal. The log of 1 is zero, so pH = pKa.

This is true regardless of whether or not it's an acid or a base... but the pKa will just be very high or very low depending on how acidic or basic it is, and thus the pH for the half-half point will change accordingly.

Thanks again

I always get molecular geometry questions wrong... is there a step by step process to figuring them out.... including bond angles? I get confused by the presence of lone pairs

For ex: XeF2 is linear.. how? What about H2O and NH3. I can never find a foolproof way of getting the geometry and bond angles right... costs me free points on PS =(

First of all you have to memorize the set of possibilities: linear, trigonal planar, tetrahedral, trigonal bypramidal, and octahedral. If you don't have them memorized or you can't picture them in your mind then you will have a hard time with these questions no matter what.

The first thing to do with a molecular geometry question is to sum up the number of valence electrons you have. Lets use H2O as an example, each hydrogen is in the first column of the periodic table and therefore contributes one valence electron, the oxygen likewise contributes six, for a grand total of 8 (1+1+6).

Next you have to make an educated guess as to what layout you expect the molecule to have. Hydrogen will always be on the edges of the molecule, so that means oxygen is probably in the middle. We would expect H2O to look something like

H-O-H

Now fill in the valence shells of each atom with lone pairs of electrons. Hydrogen can only hold 2 in it's shell, which each already have from their bonds. Oxygen has 4 electrons from the bonds so it needs 4 more, which is two lone pairs. Draw those in as dots above and below it. Now count up the number of electrons you have in the diagram and compare it to the number you calculated in the beginning of the problem. We calculated 8 in the beginning, and we now count 8 in our figure (4 from the bonds, 4 from the lone pairs), so we are good to go. If, however, we had a molecule where this was not the case you would have to play around by inserting double bonds or over-filling the octets of some atoms (only acceptable for atoms with d-orbitals) until you matched electron counts.

Next, recognize that the oxygen has 4 groups (or "ligands") around it, the two hydrogens and the two lone pairs. Any atom with 4 groups around it is going to be tetrahedral (this is where the memorization comes into play) and therefore the bond angles will be close to 109.5 degrees.

The oxygen itself is tetrahedral, but the entire water molecule is said to be "bent". Picture the tetrahedral oxygen in your mind, two of the ligands off of it are hydrogens and two are lone pairs of electrons. We care about the hydrogens, and this is difficult to show without a model but the two hydrogens and the oxygen form a kind of V shape, which is called "bent." The H-O-H bond angle in this case turns out to be slightly less than 109.5 degrees because of the repulsion from the two lone pairs of electrons, but I'm not sure if you will need to know that for the MCAT.

XeF2 is harder than H20, but you can use the exact same approach. It turns out to be trigonal bipyramidal with the fluorines occupying the two tips of the molecule and lone pairs of electrons in the three center positions. The fluorines are then in a linear conformation with the central xenon atom.

So you basically have to know the different types of shapes, draw the molecule, figure out the bond angle... but for the actual shape.. you have to kinda "subtract" the lone pairs and see what it looks like?

Thanks.. I think I just didn't have a good picture of what tetrahedral or trigonal bypyrimidal looked like ... so I was never able to decide between linear and bent, for example.

axp107 said:

for the actual shape.. you have to kinda "subtract" the lone pairs and see what it looks like?

Click to expand...

Yeah exactly. For these it's really helpful to be able to picture the molecules in your mind, although I suppose it would be possible to do it just by memorization.

If the reaction was:

A3+ + H2O <-> A + OH- + H+ (fast)
A3+ + OH- <-> AOH2+ (slow)
H+ + AOH2+ + B2+ <-> A2+ + H2O + B3+ (slow)

If I were to find the rate of this reaction, I would use slow since it's rate-determining. So

rate = k[A3+] [OH-]
Then we would substitute [OH-] using the rate of OH- formation from the previous (fast) reaction.

My question is that we can't know the order of the overall reaction until we prove through experiments. So how can we say that the rate of the reaction is indeed rate = k[A3+] [OH-] and rate of formation of OH- = k[A3+]? I feel that we can't know if it's [A3+]^2 or [A3+] just from the reaction above without any experimental data. Can anybody let me know what's wrong with my thought?

I've been attempting to put together a list of thermo/kinetic quantities in order to be prepared for the many q's where they ask the effect of a thermo change on a kinetic attribute or vice versa and was hoping to see if others can add to it.

Thermo
Activation Energy
K(sp)
Enthalpy
Gibbs
Entropy
Heat of Salvation


Kinetic
Temp
Conc
Pressure
Order
Kinetic Energy

anks106 said:

Heat of Salvation

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Isnt that zero for all cases where salvation occurs? It increases without bound when there is no salvation.

edit.

I tend to get confused when I see compounds like this:

ZnCl3.OH

what does the "." even mean.. is it a covalent bond? How would reactions occur with such a compound?

What if we were asked to find the oxidation number of Zn, how would we go about doing it?

axp107 said:

I tend to get confused when I see compounds like this:

ZnCl3.OH

what does the "." even mean.. is it a covalent bond? How would reactions occur with such a compound?

What if we were asked to find the oxidation number of Zn, how would we go about doing it?

Click to expand...

I assume that molecule is essentially Zn with three Cl in polar covalent bonds around it, with a charge of +, and OH-. Zn has a very low electronegativity so its bonds with both Cl and O are polar covalent or ionic. The period is kinda weird - I've never seen that before.

For oxidation numbers, just do Zn last. O is 1- and each Cl is 1-, so that's 4-. H is 1+, so for the compound to be neutral (though technically it splits into its ions readily), Zn must be 3+.

EmiliaC said:

I assume that molecule is essentially Zn with three Cl in polar covalent bonds around it, with a charge of +, and OH-. Zn has a very low electronegativity so its bonds with both Cl and O are polar covalent or ionic. The period is kinda weird - I've never seen that before.

For oxidation numbers, just do Zn last. O is 1- and each Cl is 1-, so that's 4-. H is 1+, so for the compound to be neutral (though technically it splits into its ions readily), Zn must be 3+.

Click to expand...

I've seem them on AAMCs before.. an ionic compound combined with an H2O or something like...

NaCl.H2O

Do you even factor the H2O in? What does it mean to have an H2O group attached?

axp107 said:

I've seem them on AAMCs before.. an ionic compound combined with an H2O or something like...

NaCl.H2O

Do you even factor the H2O in? What does it mean to have an H2O group attached?

Click to expand...

If it's H20 attached, it means the compound is anhydrous (saturated with a certain number of water molecules for every molecule of the compound of interest).

EmiliaC said:

If it's H20 attached, it means the compound is anhydrous (saturated with a certain number of water molecules for every molecule of the compound of interest).

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Actually, if the compound contains water molecules, it's said to be a hydrate. If there are no water molecules, it's considered anhydrous ("an" = not).

When figuring out reduction potentials, say you have the reaction:

2 Xx + 2 Yy --> 2Xy + 2 Yx E(cell)=2 V

To find the E(cell) of the reaction Xx + Yy --> Xy + Yx, do you just divide the other E(cell) by 2 since it's only half of the moles? Would the E(cell) be 1 V?

Sorry if that was confusing...

this question got lost in the general chem thread, so i'm reposting hopefully w/ at least 1 response!!!

i have a quick question on beta decay (conversion of neutron into proton, electron, and antineutrino). when the electron is ejected, wouldn't that cause the daughter isotope to have a + charge?

e.g., (A/Z X) 14/6 C -> 14/7 N + 0/-1 e- this balances out. but then the 0/-1 e- is ejected, so i would assume that 14/7 N would become 14/7 N+

would this happen? why or why not? thanks in advance...

what is the oxidation state of Mn in MnO2OH?

O2=-4
OH=-1
Mn=+5

On one of Kaplan's flashcards, it states that a real gas will have less then expected volume at low temperatures. Why is that? Is it because when the temperature is low, the K.E. of the molecules will decrease and intermolecular attractice forces act on the gases?? Then why is the volume higher then expected from an ideal gas when at very high pressure? Is it because of the forced boundaries? Thanks

HristosKaran said:

On one of Kaplan's flashcards, it states that a real gas will have less then expected volume at low temperatures. Why is that? Is it because when the temperature is low, the K.E. of the molecules will decrease and intermolecular attractice forces act on the gases??

Click to expand...

Yup, thats right. As the K.E. decreases, the molecules slow down. This allows intermolecular forces to become more significant.

HristosKaran said:

Then why is the volume higher then expected from an ideal gas when at very high pressure? Is it because of the forced boundaries? Thanks

Click to expand...

Yes, its because of the "forced boundaries", but I'll give you the Kaplan explanation, followed by an analogy to drive the point home (which may end up confusing you. If this is the case, ignore it).

Kaplan explanation:
"The size of the particles becomes relatively large compared to the distance between them."

Analogy:
Imagine two magnets with an opposing force between them. At first, if you push the two together, the distance between them will decrease. But once they start touching each other, they can't be compressed anymore. No matter how much you push them together, they can't move any more because they are already touching each other. Even at very high pressures, they absolutely won't compress anymore.

The PV=nRT equation assumes an inverse relation between P and V. But this breaks down at high pressures, because the pressure will increase, but the volume will stay relatively constant.

Hope that helps.

elayem said:

this question got lost in the general chem thread, so i'm reposting hopefully w/ at least 1 response!!!

i have a quick question on beta decay (conversion of neutron into proton, electron, and antineutrino). when the electron is ejected, wouldn't that cause the daughter isotope to have a + charge?

e.g., (A/Z X) 14/6 C -> 14/7 N + 0/-1 e- this balances out. but then the 0/-1 e- is ejected, so i would assume that 14/7 N would become 14/7 N+

would this happen? why or why not? thanks in advance...

Click to expand...

Yup, this is what would happen.

The neutron turns into a proton and electron. But then the electron gets ejected, leaving a +1 charge on the daughter.

chad5871 said:

When figuring out reduction potentials, say you have the reaction:

2 Xx + 2 Yy --> 2Xy + 2 Yx E(cell)=2 V

To find the E(cell) of the reaction Xx + Yy --> Xy + Yx, do you just divide the other E(cell) by 2 since it's only half of the moles? Would the E(cell) be 1 V?

Sorry if that was confusing...

Click to expand...

No, you don't multiply by the number of moles. It is still the same.

E is an intensive property, so it remains the same, regardless of the number of moles.

i hope i'm posting this in the right place, there seem to be a lot of q&a areas.

the bronsted lowry definition of a base is something that accepts a proton.

i'm hoping someone can help me understand how a strong base like NaOH accepts a proton in water. i believe the equation looks like:

NaOH + H2O > OH- + Na+ + H2O

i suppose i could have just as easily left out the H2O, which is kind of the point.

isn't this just the dissociation of the NaOH molecule?

in what sense is the NaOH accepting a proton?

many thanks for any help!

CremasterFlash said:

i hope i'm posting this in the right place, there seem to be a lot of q&a areas.

the bronsted lowry definition of a base is something that accepts a proton.

i'm hoping someone can help me understand how a strong base like NaOH accepts a proton in water. i believe the equation looks like:

NaOH + H2O > OH- + Na+ + H2O

i suppose i could have just as easily left out the H2O, which is kind of the point.

isn't this just the dissociation of the NaOH molecule?

in what sense is the NaOH accepting a proton?

many thanks for any help!

Click to expand...

It might be a proton transfer.

NaOH + H2O --> OH- + Na+ + H2O

Ok, but which water? Let's radioactively label the oxygen on the NaOH.

NaOH + H2O --> OH- + Na+ + H2O

So in this case, we have the OH- from the NaOH pulling off the H+ from the water in solution, and then the water in solution is left just as OH-. Therefore, we did have a proton transfer. But really, it doesn't matter a whole lot in this specific instance, since either molecule is equally likely to have the H+ on it. That is to say, the OH- on the right side of the reaction could just pick up the H+ from the H2O.

Hope this wasn't confusing.

so the OH- resulting from the NaOH disassociation immediately strips a proton off a nearby H2O, resulting in a new OH- which then does the same thing immediately. so the concentration of OH- stays the same but its composition is changing constantly (in terms of the specific OH-s that constitute the [OH]). at least i think that's what you're saying.

so a weak base, say NH3, receives a H+ from H2O, creating NH4+ and OH- and this OH- does the same thing, taking protons from H2O and creating new OHs.

so in what way is the distinction between an Arrhenius base and a Bronsted Lowry base important?

seems they both wind up with OH- in solution.

CremasterFlash said:

so the OH- resulting from the NaOH disassociation immediately strips a proton off a nearby H2O, resulting in a new OH- which then does the same thing immediately. so the concentration of OH- stays the same but its composition is changing constantly (in terms of the specific OH-s that constitute the [OH]). at least i think that's what you're saying.

so a weak base, say NH3, receives a H+ from H2O, creating NH4+ and OH- and this OH- does the same thing, taking protons from H2O and creating new OHs.

so in what way is the distinction between an Arrhenius base and a Bronsted Lowry base important?

seems they both wind up with OH- in solution.

Click to expand...

Well the thing is, NaOH is pretty much never NaOH in water. It's a strong base, which means it completely dissociates. So there should be no "NaOH" in water. When they write NaOH(aq) they really mean that there are Na+ and OH- ions in solution. The only time when NaOH will be NaOH is if there is no water around-- you may have seen this before, they are little white pellets.

"...resulting in a new OH- which then does the same thing immediately. so the concentration of OH- stays the same but its composition is changing constantly"

Yes, exactly... we have an equilibrium so the concentration of [OH-] is the same, but each individual OH- may take up a H+ from H2O, leaving the H2O as OH- and making the OH- into H2O. It's an equilibrium-- remember in equilibrium the reaction doesn't stop, just the net direction of the reaction doesn't change.

Yes, any OH- can take a H+ from H2O and make the H2O into OH-. The reaction just looks funny because it looks like nothing is happening in the NaOH reaction.

The distinctions may be important-- review your general chemistry text for definitions. If any of the definitions are confusing, type them up here and I'll explain them.

Can someone go over the "refrigerator" rules for changes in volume vs pressure? Examkrackers vaguely attempted to explain it, but I really couldn't follow it.

So when a gas expands, work is done by the system, and the temperature is increased? While when a compressor reduces the volume, work is done on the system, and the temperature decreases? I might have that backwords...thanks
Deals with W=P dV

HristosKaran said:

Can someone go over the "refrigerator" rules for changes in volume vs pressure? Examkrackers vaguely attempted to explain it, but I really couldn't follow it.

So when a gas expands, work is done by the system, and the temperature is increased? While when a compressor reduces the volume, work is done on the system, and the temperature decreases? I might have that backwords...thanks
Deals with W=P dV

Click to expand...

W=PdV

When dV is positive, work is done by the system.
When dV is negative, work is done to the system.

Does it take energy, or release energy to squeeze a balloon?

BlackSails said:

W=PdV

When dV is positive, work is done by the system.
When dV is negative, work is done to the system.

Does it take energy, or release energy to squeeze a balloon?

Click to expand...

It takes energy to squeeze a balloon. because dv is negative and hence work must be done on the system to achieve this feat.

slyloxy said:

It takes energy to squeeze a balloon. because dv is negative and hence work must be done on the system to achieve this feat.

Click to expand...

I understand that, but how does it relate to temperature?

Edited: Problem posted in wrong thread

Posted in wrong thread.

HristosKaran said:

I understand that, but how does it relate to temperature?

Click to expand...

When a gas expands, it uses work (energy) to do so. As the gas uses energy, the temperature decreases. (The energy of atoms is measured by how fast the atoms are moving/hitting each other. Temperature is directly proportional to energy. So when energy decreases, temperature decreases.)

Example: Take one of those keyboard dust cleaner sprays with tetrafluoroethane, and spray it for a few seconds. Notice how the can gets cold? The compressed gas inside expands as it gets to the nozzle, using its own energy. This is why the temperature drops rapidly.


When a gas decreases in volume, work gets done on it by the surroundings. The gas recieves energy in the process, which causes the temperature to rise.

HristosKaran said:

A uniform rod 2 m in length with a mass of 4 kg is held at an angle &#952; (between 0° and 90°) to a frictionless tabletop by a string pulling straight up 0.5 m from the upper end of the rod. What is the tension in the string?

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It's been a while since I took general chemistry, but I'm pretty sure we didn't have any tension problems in that class.

Thanks for the explanation!

What is the molar solubility of AgBr in a 0.10 M solution of AgNO3? (Ksp AgBr = 5.0 × 10–13)
0.10 mole/liter
1.0 x 10-11 mole/liter
5.0 x 10-12 mole/liter
5.0 x 10-13 mole/liter

Now the answer is C...but how can the molar solubility increase if its saturated solubility is at 5x10 -13?? Seems kind of counter intuitive to me..

HristosKaran said:

What is the molar solubility of AgBr in a 0.10 M solution of AgNO3? (Ksp AgBr = 5.0 × 10–13)
0.10 mole/liter
1.0 x 10-11 mole/liter
5.0 x 10-12 mole/liter
5.0 x 10-13 mole/liter

Now the answer is C...but how can the molar solubility increase if its saturated solubility is at 5x10 -13?? Seems kind of counter intuitive to me..

Click to expand...

I think you've confused the concept of the common ion effect.

You said "how can the molar solubility increase if its saturated solubility is at 5x10 -13??" You are comparing the molar solubility with the Ksp, which can't be done. The common ion effect should be used to compare the molar solubility of a salt in different solutions.

The common ion effect causes a decrease in solubility of the salt in the AgNO3 solution. In other words, the solubility of the AgBr salt will be lower in AgNO3 than in water (or any other solution without Ag or Br ions), which does not have any Ag ions floating around.

The common ion effect's decrease in solubility is in relation to other solutions.



Here is a mathematical way of looking at it:

Ksp = [Ag][Br]
Molar Solubility = [Br]
Since the [Ag] = .1 M in the solution, and the Ksp is constant, the [Br] (molar solubility) must be greater than the Ksp by 10x.


Note: After typing all this up, I read it over, and it kinda sounds confusing. Let me know if you need me to clear anything up.

^^^

That makes sense, and I was heading in the right direction b/c I had common ion effect in mind, but I think I'm having a problem with the math. Let's see...

Ksp=[Ag][Br]

Since there is already .1M of Ag in solution...

5X10^-13 = (X+.1)(X)

so...x^2 (.1X)= 5X10^-13..but since the molar solubility of [Ag][Br] is very small, compared to whats already in solution, you just divide by .1. It makes sense now that I have done the problem, but its not really normal to just "forget" about the x's in the equation, and I think that's what messed me up.

What's the difference in how you calculate electric potential for these?

What the difference between these two equations:
EMF = E(cathode) - E(anode)
EMF = E(reduction) + E(oxidation)

What if an electrolytic cell is discharging would it act as a galvanic cell?

How do u guys approach these sorta questions in general? I can't seem to wrap my head around it

aclementine said:

What's the difference in how you calculate electric potential for these?

What the difference between these two equations:
EMF = E(cathode) - E(anode)
EMF = E(reduction) + E(oxidation)

What if an electrolytic cell is discharging would it act as a galvanic cell?

How do u guys approach these sorta questions in general? I can't seem to wrap my head around it

Click to expand...

I would really like to know as well...it seems every passage problem uses the reduction potentials in a different way. In exam krackers they have EMF = E (anode) - E (cathode)..can someone please clarify.

aclementine said:

What's the difference in how you calculate electric potential for these?

What the difference between these two equations:
EMF = E(cathode) - E(anode)
EMF = E(reduction) + E(oxidation)

What if an electrolytic cell is discharging would it act as a galvanic cell?

How do u guys approach these sorta questions in general? I can't seem to wrap my head around it

Click to expand...

i think you need to give us more information about your questions. The general rule is EMF = E(reduction) + E(oxidation). What was the question in which the EMF = E(cathode) - E(anode) formula was applied?

I have a question regarding chemical equilibria and why solids and liquids are not calculated into them

For an equation CaCO3(s) <---> CaO(s) + CO2(g)

Kc would equal = [CO2]

and the solids would not be factored in

My book only says since they are solids, molar concentrations are constant and that is why they are not calculated into Kc

I am not understanding the concept behind this though, a conceptual explanation would be greatly appreciated

Thanks

What volume of HCL was added if 20 mL of 1M NaOH is titrated with 1M HCL to produce a pH of 2?

The answer is 20.4 mL acid.

From what i understand is that to produce pH2 means that it already past the equivalence point (where 20ml HCL was used).
So pH=2 means that [H+]=0.01M. And i have no idea how to incorporate the [H+] too.
PLEASE HELP!!!!
Thanks in advance

well im too tired to answer the question but you can look at the question differently...how many mL of 1M HCL do you need to add to 40 mL of water to make a solution of pH 2

actually i almost forgot all of gchem its been a long time, but we know we want to dilute that 1M HCL into .01 M so 100 times so divide that 40 mL by 99 so that other 1% is the HCl you get 0.404 so the volume of that total solution is 40.4 the HCl being 20.4. if u wanted an equation i forgot all of them sorry

I might be a little off but I think the equation is:

i (sub a) x M of acid x L of acid = i (sub b) x M of base x L of base.


the i sub a and sub be are constants, M equals the molarity of acid or base, and liters equals how many girls you slept with freshman year. Just kidding! Liters equals liters.

ucla2134 said:

What volume of HCL was added if 20 mL of 1M NaOH is titrated with 1M HCL to produce a pH of 2?

The answer is 20.4 mL acid.

From what i understand is that to produce pH2 means that it already past the equivalence point (where 20ml HCL was used).
So pH=2 means that [H+]=0.01M. And i have no idea how to incorporate the [H+] too.
PLEASE HELP!!!!
Thanks in advance

Click to expand...

First:

20mL x 1M NaOH = 20 millimoles


Now, to neutralize the base, add HCl:

20mL x 1M HCl = 20 millimoles


The solution is now neutral. The total volume is now 40mL.

Next, the pH has to be adjusted to 2.

pH = -log([H+])
2 = -log([H+])
10^-2 = [H+]
[H+] = .01 M

Thus, the final concentration of the [H+] is going to be .01M.

.01 M = (1M HCl)(x) / (x + 40mL)
.01x + .4 = x
.99x = .4
x = .40mL

So the total amount of HCl that needs to be added is:

20mL to neutralize the base
0.40mL to adjust the pH

Total HCl added: 20.4mL