[CharlieTran]

Hey guys

I'm currently doing high school chemistry at university and it's really hard. >.<
But I'm willing to learn so if you guys can, can you help me with these equations? It's been bugging me for like 3 hours (per question). My final test is on in about 3 weeks.

Please show working out so I can learn how to do them myself

1. What volume of 0.100 M sodium hydroxide solution is required to react completely with 20.0 mL of 0.19 M sulfuric acid solution to reach the end point?

2.  What mass of insoluble calcium fluoride will precipitate when a solution containing excess sodium fluoride is added to a solution containing 18.2 g of calcium chloride?

3. 4.9 L of hydrogen chloride gas is dissolved in water and made up to 1 L of solution in a volumetric flask. What is the molar concentration of chloride ion in the resultant solution?

4. 20.0 mL 0.063 M sodium hydroxide is added to 20.0 mL 0.063 M nitric acid solution, and the solution is evaporated to dryness. What mass of solid is obtained?

Any of these questions would be helpful

Many thanks in advance

[LQ43]

Start by writing balanced equations for the reactions. A review of nomenclature might help if this is difficult at first

[Astrokel]

Hey, i'll give you the rough idea using qn 1.

1) You've to write out the equation of the reaction first and you will realize the stoichiometric coefficient of NaOH to H2SO4 is 2:1.

Therefore, for every 1 mole of H2SO4, you need 2 moles of NaOH for complete neutralisation.

You have the concentration and volume of H2SO4, you can use n=cv to find the amount of H2SO4 used.

Since you know the stoichiometric ratio, and you can calculate the amount of NaOH required, therefore you can also calculate the volume using v=n/c

2) Same old step, write out the equation, get the stoichiometric coefficient of CaCl₂ and CaF₂.

In this question you have to use, n = mass/Mr formula

3) Write out the equation, note, the total volume after the dissolved.

4) Usually, you'd have to work out which is the limited reagent, however in this case it is not required. Write out the equation as usual and find the stoichiometric coefficient of product required with respect to NaOH and HNO3. After this it is simple.

[CharlieTran]

thanks Astrokel but I'm still sorta suck

So the equation is 2NaOH + H2SO4 = Na2SO4 + 2H20

So 1 mole of H2So4 reacts with 2NaOH to give 0.38M

So V(l) = C (mol)/N (moles)

 = 0.38M/0.1M

= 3.8

I think I got this wrong because it doesn't look right =/

[Borek]

So the equation is 2NaOH + H2SO4 = Na2SO4 + 2H20

So far so good.

So 1 mole of H2So4 reacts with 2NaOH to give 0.38M

Why? You've lost me here.

[Astrokel]

thanks Astrokel but I'm still sorta suck

So the equation is 2NaOH + H2SO4 = Na2SO4 + 2H20

So 1 mole of H2So4 reacts with 2NaOH to give 0.38M

So V(l) = C (mol)/N (moles)

 = 0.38M/0.1M

= 3.8

I think I got this wrong because it doesn't look right =/

Hey! how did you get 0.38?

n=cv

n stands for amount in moles,
c is concentration in mol dm^-3, or usually denoted by M
v is volume in dm³

check your formula above,

for example, to get the amount of H2SO4 used,

n = (0.19)(20.0/1000)

Stoichiometric ratio works in this way,

1 mole of H2SO4 required 2 moles of NaOH for complete neutralization.
0.5 mole of H2S04 required 1 mole of NaOH for complete neutralization.
5 moles of H2SO4 required 10 moles of NaOH for complete neutralization.

so after you get moles of H2SO4 used, you can simply get the amount of NaOH required by ratio.

With that, you can find the volume required.

[CharlieTran]

Isn't it because

2NaOH + H2SO4 = Na2SO4 + 2H20

2 Moles + 1Mole = 1Mole    + 2Mole

And since we want to find the NaOH the reaction is

2 : 1?

since the mole of H2SO4 was 0.19, so mole reacted = 0.19M x 2 = 0.38?

[CharlieTran]

Wait, I got it now! yea, i think i didn't do the N=CV thing properly

Now 3 more questions to go!

[Astrokel]

Isn't it because

2NaOH + H2SO4 = Na2SO4 + 2H20

2 Moles + 1Mole = 1Mole    + 2Mole

And since we want to find the NaOH the reaction is

2 : 1?

since the mole of H2SO4 was 0.19, so mole reacted = 0.19M x 2 = 0.38?

Yes! your concept is right, just your numbers is wrong.

Wait, I got it now! yea, i think i didn't do the N=CV thing properly

Now 3 more questions to go!

ok! 

[mnakhla]

for the first one just use the formula M1V1=M2V2
which is molarity of the first times the volume which is x in this case for the naoh side and molarity 2 times the volume of sulfuric acid... and adjust the equation to take care of the the 2 to 1 acid to base ratio by adding a 2 to the appropriate side


for the 4.9L of HCL...just divide 4.9L (which is the amount of volume an ideal gas occupies...im assuming you are using ideal gas laws. by 22.41L and you will get the amount in moles of hcl in the solution and since its a 1L solution that number will also be your molarity of solution

and the last question, since the naoh and the nitric acid are in equal molar ratios they will react completely to an equilibrium of the previously mentioned species and NaNO₃ and H₂O but since you are taking the water out by vaporization you are shifting the equilibrum too sodium nitrate so the other species are negligable...meaning you just need to calculate the molar weight of sodium nitrate and multiply it times .063 moles and times the 20ml volume of the substance to get grams.

[Borek]

for the first one just use the formula M1V1=M2V2
which is molarity of the first times the volume which is x in this case for the naoh side and molarity 2 times the volume of sulfuric acid... and adjust the equation to take care of the the 2 to 1 acid to base ratio by adding a 2 to the appropriate side

That's not the best approach - even if it gives correct reasult, most students remember only formula and they blindly use it to solve any titration questions. It is better to start with reaction equation and go through old fashioned stoichiometry.

for the 4.9L of HCL...just divide 22.1 (which is the amount of volume an ideal gas occupies...im assuming you are using ideal gas laws. by 4.9L and you will get the amount in moles of hcl in the solution and since its a 1L solution that number will also be your molarity of solution

Unfortunately - no. 22.1/4.9 is not number of moles of HCl at STP. Neither 22.1 nor fraction you propose are correct.

and the last question, since the naoh and the nitric acid are in equal molar ratios they will react completely to an equilibrium of the previously mentioned species and NaNO₃ and H₂O but since you are taking the water out by vaporization you are shifting the equilibrum too sodium nitrate so the other species are negligable...meaning you just need to calculate the molar weight of sodium nitrate and multiply it times .063 moles of the substance to get grams.

I am afraid you are wrong again. First of all - you forgot about the solution volume. Second, your analysis about equilibrium is only dimming the picture. This is a simple stichiometric question, that should be started with reaction equation.

[mnakhla]

I dont know about the easiest approach i was just giving one that wasnt mentioned , the only one actually that i was ever taught in highschool and college the other equation previously mentioned i had never encountered before

but, i had just looked it up 22.1liters for an ideal gas.... i remeber using it in highschool?? did i just state the wrong number as i believe i am slightly dislecsic (sp?) and easily swich numbers around....i just looked it up again an ideal gas at stp occupies 22.41 liters courses.cm.utexas.edu/jbrodbelt/ch301/2007Fall/chapter5_ppt.ppt it is located somewhere within this power point...sorry for the vagueness

and the last one you are right i forgot to menation that he needs to multiply by the 20mls

[Borek]

22.41 is OK. But there is much larger problem - you have suggested 22.41/4.9 as number of moles. That'll mean the smaller the volume, the more substance you have - which is obviously incorrect...

[mnakhla]

ehh? thats not the post i just posted? but yes i did swich the numbers.. usually that happens unless i physically write it down...i really need to pay attention to my posts so i dont confuse ppl or give wrong answers because although it makes sense what im doing in my head i tend to make it come out all wrong

i also need sleep and for some one to answer my post about sulfuric acid synthesis in the citizen chemist section

[mnakhla]

AHA PROBLEM SOLVED INCORRECT POST MODIFIED!