[NBuro]

Hello,

Firstly, I have to say I'm new here. Also, I'm a french-speaker, so it's possible that I make some mistakes... but anyways, here's my problem :

I have to choose between 3 HCl concentrations (1M, 0.1M and 0.01M) to represent a stomach that hurts (so I chose 0.1M HCl). That's okay, we all understand here.

The 0.1M HCl has a pH of 1 and a normal stomach has a pH of 2.

I got some CaCO3 powder here and I want to put some into my 0.1M HCl (pH 1). I want the reaction to become CaCl2 of pH 2 (we don't mind about the H20 and the CO2).

I also have to calculate the speed reaction (I don't know if the french translation is the same... but the letter is V) of the reaction.

I just want to know how much CaCO3 (normal chalk) is needed. So how much CaCO3 grams per how much 0.1M HCl mL is needed to get a solution of CaCl2 of pH 2.


It may be hard to understand because of my mistakes... anyway, tell me if something is wrong, I'll explain you.

Thks...


oh! And here's a small summary of what I want :

X(mL) of 0.1M HCl (pH1) +  Y(grams) of CaCO3 (normal board chalk) --> CaCl2 (pH2) + H20 (don't mind) + CO2 (don't mind)
X = ?, Y = ?
What is the speed reaction (V)?

[Borek]

Not enough information to calculate reaction speed. As for the rest - it is simple stoichiometry. Start with the balanced reaction equation.

[NBuro]

OK!

But... I'm not good in stochiometry... could you help me please?

I don't know the concentration of normal chalk (CaCO3) and also what would be CaCl2's one for pH 2

I'm not really good in chemistry, even if I look up in my books, I can't figure out how to do that...

 

thks

[Borek]

I don't know the concentration of normal chalk (CaCO3) and also what would be CaCl2's one for pH 2

There will be no solution - suspension at best. CaCO₃ is insoluble, so you have to treat it as solid. CaCl₂ concentration doesn't matter, as it will be defined by the reagent amounts - thing that you have to calculate.

Start with the balanced reaction equation.

[NBuro]

CaCO₃ + 2 HCl= CaCl₂ + H₂O + C0₂

Sorry... that's all I know about it
 :hrmm:

And how will I know that my CaCl₂ will have a pH of 2?

[Borek]

CaCO₃ + 2 HCl= CaCl₂ + H₂O + C0₂

Sorry... that's all I know about it

But that's half of the correct solution

And how will I know that my CaCl₂ will have a pH of 2?

To have pH=2 you must left (after neutralization) some hydrochloric acid. You know what was the starting concentration, you know what should be final concentration (both defined through pH value) - you can calculate how much acid needs to react. You have the properly balanced reaction equation above - so you have all necessary information at hand.

Do the calculations per 100mL of stomach juice, as they can't be done without known volume.

[NBuro]

so I need 0.01M of CaCl₂?

Okay, let's try that :

CaCO3 = 100 g/mol
HCl = 36 g/mol
CaCl2 = 110 g/mol
H2O = 18 g/mol
CO2 = 44 g/mol

CaCO3 + 2 HCl = CaCl2 + H2O + CO2

100 mL of HCl (0.1 mol/L) --> 0.01 mol  --> 0.36g of HCl

Let's hope I'm getting in the right way, now :

?.?? mL of CaCl2 (0.01 mol/L) --> ?.?? mol --> ?.??g of CaCl2

  what now?

[Borek]

so I need 0.01M of CaCl₂?

No.

You start with 0.1M HCl, you need to neutralize it to left 0.01M HCl - so 0.09M must react. But 1 mole of HCl produces only 1/2 mole of CaCl₂ (see reaction equation).

However, you don't need to calculate CaCl₂ concentration. You have to calculate how much CaCO₃ is necessary to neutralize 0.09M HCl solution (per 100 mL, as otherwise the question doesn't ake sense). These amounts (CaCl₂ & CaCO₃) will be identical - same number of moles (see reaction equation), but for obvious reasons different masses.

[NBuro]

CaCO3 + 2 HCl = CaCl2 + H2O + CO2

100 mL (0.1 L) of HCl (0.1 mol/L) --> 0.01 mol

so I have 0.01 mol of HCl here?

but I only need 0.009 mol to react... and if 1 mol of HCl produces 1/2 of CaCl2, then I'll get 0.0045mol of it?

0.009 mol of 2 HCl = 0.324g --> 0.018g of 2 H and 0.306g of 2 Cl

Okay. 0.0045mol of CaCl2 = 0.475g --> Xg of Ca and 0.306g of Cl2, so X = 0.169

No way... I'm sure I'm all wrong here...

so...

CaCO3 + 2 HCl = CaCl2 + H2O + CO2

so CaCO3 = 0.169 + ?

...nah, I'm still sure I'm out of the track.... am I?

[Borek]

CaCO3 + 2 HCl = CaCl2 + H2O + CO2

100 mL (0.1 L) of HCl (0.1 mol/L) --> 0.01 mol

so I have 0.01 mol of HCl here?

Si, Señor

but I only need 0.009 mol to react... and if 1 mol of HCl produces 1/2 of CaCl2, then I'll get 0.0045mol of it?

So far, so good.

0.009 mol of 2 HCl = 0.324g --> 0.018g of 2 H and 0.306g of 2 Cl

Something is wrong, besides, you don't need these masses.

Okay. 0.0045mol of CaCl2 = 0.475g --> Xg of Ca and 0.306g of Cl2, so X = 0.169

No way... I'm sure I'm all wrong here...

You are right - you are wrong. There is no need for all these calculations (and the results are incorrect for some reason).

CaCO3 + 2 HCl = CaCl2 + H2O + CO2

You have calculated correctly that 0.009 mol of HCl must react (per 100 mL), you have calculated correctly that 0.0045 mole of CaCl₂ will be produced. How many moles of CaCO₃ must react to produce 0.0045 mole of CaCl₂? (Note that you could calculate amount of CaCO₃ the same way you calculated amount of CaCl₂, just looking at the reaction equation).

Don't go for masses, start calculating moles, convert to mass at final stage.

[NBuro]

It's okay now! I showed it to some friends at school and we did it together!

0.0045 CaCO3 + 0.009 HCl = 0.0045 CaCl2 + 0.0045 H2O + 0.0045 CO2

That's it... and I only need the mass of CaCO3

I get 0.45g of CaCO3 per 100mL of 0.009 HCL...

I made the experimentation today, and it worked well. My final pH was 1.8 (surely because I misweighted the CaCO3), but it worked.

So I wanted to thank you Borek, for your help    You helped me to pass through it, plus : now I understand more!

[Borek]

You are welcome