Neutralization Question

[HopefulOncoDoc]

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Hey guys sorry if this question is a bit easy but I just had a quick question as I'm having trouble understand a specific part of the solution. So the question is...

How many mL's of 0.60 M HCl are required to neutralize 3.0 grams CaCO3?
A. 50 mL
B. 100 mL
C. 200 mL
D. 300 mL

I put A initially but the answer was B. Anyways on the solution they put

moles of OH- = 2 x moles CO3 2- = (M H+)(V H+)

I'm confused as to why we are multiplying 2 x mol CO3 2- . Shouldn't we be multiplying 2 x (M H+)(V H+) since according to the balanced equation

CaCO3 (s) + 2HCl (aq) <---> CaCl2 (aq) + CO2 (g) + H2O (l)

for every CaCO3 we react with 2HCl? Thanks a lot everyone.

 

[cartman1980]

shouldn't the equation be

2CaCO3 (s) + 4HCl (aq) <---> 2CaCl2 (aq) + 2H2CO3

 

[loveoforganic]

shouldn't the equation be

2CaCO3 (s) + 4HCl (aq) <---> 2CaCl2 (aq) + 2H2CO3

Every one of your coefficients is divisible by 2


OP, the reaction equation Cartman wrote is correct. Carbonate (CO3--) is a dianion, so it gets protonated twice to form carbonic acid (H2CO3), which decomposes to CO2 and H2O

 

[cartman1980]

pen before the head... 🙂 sorry, but hopefully it answers the question..

 

[HopefulOncoDoc]

Hey guys thanks for the response. Sorry I feel so stupid...I know CO32- is going to get protonated twice so shouldnt we need 2 equivalents of HCl to do that? And so knowing that I'm just confused as to why shouldn't it be

2 (M H+)(V H+) = moles CO3 2-

 

[Rabolisk]

For every mole of carbonate, you need two moles of HCl. So if I tell you the value of [CO3^2-], you need to multiply that by two to get [H+]. Your equation implies that for every mole of HCl, you have two moles of carbonate.

 

[HopefulOncoDoc]

For every mole of carbonate, you need two moles of HCl. So if I tell you the value of [CO3^2-], you need to multiply that by two to get [H+]. Your equation implies that for every mole of HCl, you have two moles of carbonate.

Ah Yes I see that makes a lot sense. Thanks a lot guys!

 

[indianjatt]

Sorry to butt in, but I'm somewhat lost.

Even though Carbonate is protonated twice, isn't the equilibrium constant very small; definitely not 100% yield, possibly a 1% of the reaction proceeds to form OH-. So why are we assuming in this case that it will be a quantitative reaction and the # moles CO32- = 2x moles OH-? I would assume it'd be a lot smaller.

Maybe i'm thinking of this wrong and it has nothing to do with ice tables or Ka values?

 

[Rabolisk]

The overall reaction will proceed to completion because the reaction between H+ and OH- will go to completion.

 

[cartman1980]

apologies to dig up an old question. but instead of using the formula (which is essentially # of moles of acid == # of moles of base -- for a neutralization reaction), i tried to solve this using a slightly inconvenient method.. but can't seem to arrive at the same answer..

100gm CaCO3 requires 73gm HCL to allow neutraliziation
3gm CaCO3 will need (3*73)/100gm HCL

0.6M HCL contains 0.6 moles HCL in 1000ml soln
therefore, (3*73/100) moles HCL will require (3*73)*1000/(100*.6)

the answer obviously isn't the same (and incorrect), but i can't seem to rationalize why 🙁

 

[loveoforganic]

nowhere have you incorporated the MW of HCl into your equations, so how are you converting from grams HCl to moles HCl

 

[cartman1980]

nowhere have you incorporated the MW of HCl into your equations, so how are you converting from grams HCl to moles HCl

HCL MW = 36.5gm
since one mole of CaCO3 needs 2 moles of HCL
100gm CaCO3 would need 2*36.5=73 gm of HCL
no?

 

[loveoforganic]

Ah, gotcha, just did a quick glance and saw that missing, will look through them more closely, just a min

therefore, (3*73/100) moles HCL will require

What're you jumping to here? Why do you need that many moles of HCl? That's how many grams you need, not moles. (3*73/100) grams HCl / (36.5 g HCl / mol) = 0.06 mol HCl / (0.6 mol HCl / L) = 100 mL

 

[cartman1980]

gahhhh....
before going along this track, i was trying to go through TBR Chem chapter on acid/bases, and boy did i feel lost.. made me wonder how the hell did i even graduate.. figured i could use a mood booster, so switched to a topic that i thought i knew.. and viola, the result was an incorrect answer...

many thanks for point out the obvious mistake.. (feeling the MCAT burnout.. .its been 4 months now)..