TBR Acids&Bases

[moose45]

---

Members don't see this ad.

Anyone know what is going on in Example 4.13 on page 252 of the Berkeley Review General Chemistry Book??

 

[BerkReviewTeach]

Anyone know what is going on in Example 4.13 on page 252 of the Berkeley Review General Chemistry Book??

I think I have a different edition, because Example 4.13 is on page 248. But it's about HCl neutralizing CaCO3, right?

The trick to this one is that the moles of acid you need are twice the moles of CaCO3, because it takes two H+ equivalents to neutralize CaO32-. 3.0 g of CaCO3, at 100 g/mole, represents 30 millimoles CaCO3. This means that you need 60 millimoles of HCl for complete neutralization.

The solution is 0.60 M HCl, so when multiplying 0.60 moles/L by 100 mL, you get the L cancelling out and are left with 60 millimoles of HCl. This means that 100 mL are needed.

Let me know if that's the same question and if that explains it.

 

[moose45]

Thanks for your reply. That is the correct example I was referring too but I am still confused because the way I am seeing it is different.

If you require 2 equivalents of hydrogen ion, then the calculations should be:

2 x 0.6M x V = (3/100) moles which I think eventually gives 25mL of hydrogen ion.

Why would my explanation not be correct?? Thanks!

 

[BerkReviewTeach]

Thanks for your reply. That is the correct example I was referring too but I am still confused because the way I am seeing it is different.

If you require 2 equivalents of hydrogen ion, then the calculations should be:

2 x 0.6M x V = (3/100) moles which I think eventually gives 25mL of hydrogen ion.

Why would my explanation not be correct?? Thanks!

That is a really common math mistake. Misplacing the factor of 2 is common with stoichiometry, reaction rates, and other things of the sort.

Let's say you had 1 mole of CaCO3. According to the equation, you would need twice as many moles of HCl as CaCO3, so you would need (2 x 1) moles of HCl total. So, when you have (3/100) moles CaCO3, then according to the stoichiometry, you need (6/100) moles HCl.

To get (6/100) moles HCl, you have 0.6M x V = 6/100, so V 1/10 L = 100 mL

 

[Will Hunting]

Thanks for your reply. That is the correct example I was referring too but I am still confused because the way I am seeing it is different.

If you require 2 equivalents of hydrogen ion, then the calculations should be:

2 x 0.6M x V = (3/100) moles which I think eventually gives 25mL of hydrogen ion.

Why would my explanation not be correct?? Thanks!

When you can, balance the equation

2HCl + CaCO3= CaCl2 + H2CO3
1:2
so, you know that the amount of H+ must be twice CO3

You're logic would have worked for this problem

HCl + NaHCO3= NaCl + H2CO3
1:1