[millerst]

Hello,

I have been attempting these questions for a bit now and I have not been able to come up with an answer. Could someone please help me with the following question?


A student was to make up a 55Ml of a 0.60 mol/L solution of HCL by diluting it from a solution of the same acid with a concentration of 1.00 mol/L.

a) What volume of the more concentrated solution would the student need?

C1= 1.00 mol/L               V1 = C2V2 / C1  = (0.60)(55)/1.00 = 33 mL
C2= 0.60 mol/L
V1= ?
V2= 55 mL

**If someone could tell me if the above is correct I'd appreciate that.


b) Concentration of the diluted acid was tested by titrating it with 0.100 mol/L solution of sodium hydroxide. A 10 mL sample of the HCL took 5.8 mL of sodium hydroxide solution to turn pink.


    i) How many moles of sodium hydroxide were used in titration?

Sodium hydroxide = NaOH    Molar Mass = 40g/mol
**I'm not sure how to do this considering they give a volume instead of a weight. Is there a special formula for this or something?


    ii) Write the balanced chemical equation of the reaction.

NaOH + HCl -> NaCl + H2O


   iii) How many moles of HCL were in the 10 mL sample.

**Once again I do not even know where to begin, some if someone could tell me where to start that would be great.


    iv)Calculate the actual concentration of the HCL that the student produced.

**Once again I do not even know where to begin, some if someone could tell me where to start that would be great.

[Yggdrasil]

A student was to make up a 55Ml of a 0.60 mol/L solution of HCL by diluting it from a solution of the same acid with a concentration of 1.00 mol/L.

a) What volume of the more concentrated solution would the student need?

C1= 1.00 mol/L               V1 = C2V2 / C1  = (0.60)(55)/1.00 = 33 mL
C2= 0.60 mol/L
V1= ?
V2= 55 mL

**If someone could tell me if the above is correct I'd appreciate that.

Correct.

b) Concentration of the diluted acid was tested by titating it with 0.100 mol/L solution of sodium hydroxide. A 10 mL sample of the HCL took 5.8 mL of sodium hydroxide solution to turn pink.


    i) How many moles of sodium hydroxide were used in tetration?

Sodium hydroxide = NaOH    Molar Mass = 40g/mol
**I'm not sure how to do this considering they give a volume instead of a weight. Is there a special formula for this or something?

How many moles of NaOH are in 5.8mL of a 0.100 mol/L solution?

ii) Write the balanced chemical equation of the reaction.

NaOH + HCl -> NaCl + H2O

Correct.

iii) How many moles of HCL were in the 10 mL sample.

**Once again I do not even know where to begin, some if someone could tell me where to start that would be great.

Assume all of the number of moles of NaOH calculated in (i) reacted with HCl.  How much HCl is needed to react with that quantity of NaOH?

iv)Calculate the actual concentration of the HCL that the student produced.

**Once again I do not even know where to begin, some if someone could tell me where to start that would be great.

This should be easy once you figure out (iii).

[millerst]

Well here are the first two...

How many moles of NaOH are in 5.8mL of a 0.100 mol/L solution?

          V      *        C         =       N
(0.058 mL NaOH)(0.100 mol) = 0.0058 mol NaOH

Assume all of the number of moles of NaOH calculated in (i) reacted with Hcl.  How much HCl is needed to react with that quantity of NaOH?

Okay, since it's a 1 to 1 ratio...

N_NaOH = N_HCl      Therefore there are 0.0058 mol HCL?

This should be easy once you figure out (iii).

C = N/V       =  0.0058 / 10 mL    =  0.00058 mol/L?

[Yggdrasil]

The way you set up your calculations is correct.  You just have an error in that V = 5.8mL = 0.0058L, not 0.058L

Also when figuring out moles / L, you should divide by 0.01L instead of 10mL.

[millerst]

Thank you so much Yggdrasil! Here is my next found of questions... I find these one's way more challenging then the previous ones.



1) a) An antacid tablet contains 0.9129g of aluminum hydroxide, Al(OH)₃. Find the number of moles of Al(OH)₃ in the tablet.

Molar Mass Al(OH)₃ = 78g/mol
0.9129g Al(OH)₃ / 78g/mol = 0.012 mol Al(OH)₃


b) Write the balanced equation of the reaction that occurs when the antacid dissolves in stomach acid (HCl). Determine the number of moles that can be neutralized.

** I'm not sure about this one, however, is it the same as the ones above with the C = N/V or is there some limiting thing because there are only certain amounts of moles that can be neutralized?


c) What volume of 0.10 mol/L of HCl can one tablet neutralize?

**This question, well I really have no idea where to begin...

[Yggdrasil]

1) a) An antacid tablet contains 0.9129g of aluminum hydroxide, Al(OH)₃. Find the number of moles of Al(OH)₃ in the tablet.

Molar Mass Al(OH)₃ = 78g/mol
0.9129g Al(OH)₃ / 78g/mol = 0.012 mol Al(OH)₃

Correct, although you should try to get the correct number of significant figures.

b) Write the balanced equation of the reaction that occurs when the antacid dissolves in stomach acid (HCl). Determine the number of moles that can be neutralized.

** I'm not sure about this one, however, is it the same as the ones above with the C = N/V or is there some limiting thing because there are only certain amounts of moles that can be neutralized?

First write the balanced chemical equation for the reaction between Al(OH)₃ and HCl.  Next, determine how many moles of HCl are neutralized per mole of Al(OH)₃ (here you are assuming HCl is in excess so Al(OH)₃ will be your limmiting reagent).  Finally, determine the number of moles of HCl neutralized by the number of moles of Al(OH)₃  that you calculated in part (a).

c) What volume of 0.10 mol/L of HCl can one tablet neutralize?

Once you get (b), try this one yourself.

[millerst]

First write the balanced chemical equation for the reaction between Al(OH)3 and HCl.  Next, determine how many moles of HCl are neutralized per mole of Al(OH)3 (here you are assuming HCl is in excess so Al(OH)3 will be your limmiting reagent).  Finally, determine the number of moles of HCl neutralized by the number of moles of Al(OH)3  that you calculated in part (a).

Al(OH)3 + 3HCl -> AlCl3 + 3 H2O

0.012 mol Al(OH)3 * 3HCL
                            1 mol Al(OH)3
= 0.036 mol HCl

Therefore a total of 0.036 moles can be neturalized?
             

[Yggdrasil]

Correct.

[millerst]

What volume of 0.10 mol/L of HCl can one tablet neutralize?

 

I've been attempting this one and I think I'm getting no where... maybe this is it? I'm not sure

V = N
     C

   = (0.012) / (0.10)   = 0.12 L

So one tablet can nutralize 0.12 L?

[enahs]

What volume of 0.10 mol/L of HCl can one tablet neutralize?

 

I've been attempting this one and I think I'm getting no where... maybe this is it? I'm not sure

V = N
     C

   = (0.012) / (0.10)   = 0.12 L

So one tablet can nutralize 0.12 L?

That is correct.

Think of it like this, if you are confused.

Concentration = moles per liter.

So, concentration * volume = moles
mole * liter = moles
liter


So, you knew you had 0.012 moles to neutralize, and a concentration of 0.1 M
0.012 moles / 0.1m = 0.12L


But, like I said, you had it correct.