Chemistry Questions...

[BAJackson16]

This is my first time to post on here. I am studying for my DAT and plan to take in august. I ran across this website, and figured I would let some people help me on these questions.
Chemistry is kicking my butt, any advice??
If you have time, it would help great to answer these questions?

1. How much 4MCa(OH)2 is needed to neutralize 300 milliliters or 3MHNO3?

2. WHat is the mass, in grams, or 245 milliliters of SO2 at STP?

3. What volume of .5M NaOH is needed to neutralize 25 ml of 2M H2SO4?

4. CaF2 is added to a .1M Ca(NO3)2 solution. At what concentration of F- will CaF2 begin to precipitate? (NOTE: the Ksp of CaF2 is 4*10-11)

5. What volume of water would be needed to dilute 50 ml of a 3M HCl solution to 1M?

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[DrTacoElf]

I'll take some (Someone correct me if you see any errors please!)


1. How much 4MCa(OH)2 is needed to neutralize 300 milliliters or 3MHNO3?


Focus on Normality here
4MCa(OH)2 is 8N
3MHNO3 is 3N

thus NV = NV
N= normality
V = volume

(3N)(300) = (8N)(xml)
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2. WHat is the mass, in grams, or 245 milliliters of SO2 at STP?

ALWAYS USE THIS FOR THESE QUESTIONS
pv=nrt
n=m/MM

Thus pv=mrt/MM
(1atm)(.245L)=(.0821)(mass)(273K)/(64)

Work that out and you have your answer
112.5 ml of 4MCa(OH)2 would be needed.
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5. What volume of water would be needed to dilute 50 ml of a 3M HCl solution to 1M?[/QUOTE]

MV = MV
M= Molarity
V = Volume

(50ml)(3M) = (xml)(1M)

150 ml total volume, thus 100ml H20 are needed .
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3. What volume of .5M NaOH is needed to neutralize 25 ml of 2M H2SO4?

Same question as 1 above

NV = NV
N= Normality
V= Volume
NaOH = .5N
H2SO4 = 4N

(.5N)(xml) = (4N)(25ml)
Thus 200ml of .5M NaOH are needed.

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4. CaF2 is added to a .1M Ca(NO3)2 solution. At what concentration of F- will CaF2 begin to precipitate? (NOTE: the Ksp of CaF2 is 4*10-11)


First know that when Ksp=IP you have preciptation, then write out the equilbrium expression
CaF2---> Ca + 2F
Then Use Ion Product
[Ca][2F]
F = 2x [Ca] thus
IP=[x][2x]^2 = 4x^3
ksp = 4 x 10^-11 = 4x^3 solve for X that gives you NORMAL solubility however you have to account for a common ion, Ca, in a [] of .1M already thus do the following.

IP=[.1][2x]^2 = 4 x 10^-11
You can say [Ca] is .1 and not .1 +x b/c .1 is SO much larger than the normal [Ca] just due to dissociation of Ca. Then just solve for X which is equal to the [Ca] and remember to divide this by 2 to get the [F] since CaF2 gives 2 mol F for every 1 mol!



Hope This Helps!

 

[BAJackson16]

thanks!

 

[DrTacoElf]

No Problem