TBR Gen Chem: Section VI Acids and Bases

[projectpat]

Sorry for the length post I wanted to be concise but the question at hand is quite simple..So I have a question about a pretty basic concept that I thought I fully understood.. neutralization reactions of acids and bases.. I have 3 contradictory (in my eyes) problems I want to show here..

In TBR Gen Chem book (2011 ed), Section IV, Example 4.12 (pg 252) reads..

How many mL of 0.40 NaOH are required to neutralize 100 mL 0.25 M H2SO4?
A) 62.5 mL
B) 80.0 mL
C) 100.0 mL
D) 125.0 mL

I understand that you set moles of acid = moles of base.. and solve.
The balanced reaction goes... H2SO4 + 2NaOH -> Na2H2SO4 + 2H2O

The math goes.. (mols OH-)(Volume OH-)=(mols H+)(Volume H+) x 2

My questions concern the placement of the "2".. I believed that you need 2 mols of OH- to neutralize the H+.. so you should multiply the OH by 2. Or do you not consider the balanced equation and just the fact that H2SO4 is DIPROTIC so you just double its moles for that reason?

2nd problem.. Example 4.13

How many mL of 0.60 M HCL are required to neutralize 3 grams CaCO3?
A)50 mL
B)100 mL
C)200 mL
D)300 mL

Balanced equation goes.. CaCO3 + 2HCl -> CaCl2 + CO2 + H2O

Math goes.. moles OH- = moles H+... 2 x (3g / 100 g/mol) = (.60M)(Vol H+)

Once again the placement of the "2" throws me off.. according to the balanced equation I believe you needed 2 equivalents of H+ to combat the base.. but the work demonstrates you double the moles of base...? Still confused..

3rd example (this one contradicts the other two)

Section IV, Acids and Bases Study Passages, Passage IX, Problem 59 (pg 275)

59. How many grams of CaCO3 are needed to neutralize 50 mL of stomach acid at pH = 2.0 completely, if the following equation represents the neutralization reaction?

CaCO3 + 2 H+ -> Ca + CO2 + H2O

A)25 mg
B)50 mg
C)100 mg
D)1.0 grams

So the [H+]= 0.01 M.. multiplied by 0.05 L = 5.0 x 10-4 moles H+
Solution reads "Only half the moles of CaCO3 are necessary to neutralize H+, bc it is a 2:1 ratio of H+ to CaCO3 in the balanced equation... So 2.5 x 10-4 moles multiplied by 100 g/mol (CaCO3 molar mass).. yields 25 milligrams.

Intuitively and according the balanced equation, I would need to half this value in order to get the correct answer.. which is the the correct way THIS problem is worked out.. the way it should be solved.

But According to the past two problems, I would DOUBLE this value...

So basically when they hell and where the hell do I double the values when necessary? Is based on the balanced equation..? Or is it based on if we're dealing with a monoprotic/diprotic acid..? If i'm not being clear about my question i'll try to rephrase or be more descriptive.. thank you!

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[circulus vitios]

My questions concern the placement of the "2".. I believed that you need 2 mols of OH- to neutralize the H+.. so you should multiply the OH by 2. Or do you not consider the balanced equation and just the fact that H2SO4 is DIPROTIC so you just double its moles for that reason?

The balanced equation has a 2 in it because it's diprotic.

Once again the placement of the "2" throws me off.. according to the balanced equation I believe you needed 2 equivalents of H+ to combat the base.. but the work demonstrates you double the moles of base...? Still confused.

Their "all in one line" solutions are confusing. I would do it like:

1. Write an equation
2. Find number of moles of base
3. Find number of moles of acid to neutralize base (going by the mol ratios)
4. Find volume of acid needed.

1. CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
2. 3 g base/(100 g/mol) = 0.03 mol base.
3. Equation says you need two mol acid per 1 mol of base, so 0.03 mol base * (2 mol acid/1 mol base) = 0.06 mol acid
4. 0.06 mol acid / (0.60 mol/liter) = 0.1 liters = 100 mL.

Intuitively and according the balanced equation, I would need to half this value in order to get the correct answer.. which is the the correct way THIS problem is worked out.. the way it should be solved.

But According to the past two problems, I would DOUBLE this value...

1. CaCO3 + 2 H+ -> Ca + CO2 + H2O
2. [H+] = 10^-2 = 0.01 mol/L * 0.05 L = 0.0005 mol acid
3. You're trying to find moles of base and your equation says you need 2 moles of acid per 1 mole of base. So divide your mol of acid by 2 (or multiply it by 1/2, however you want to look at it.) 0.0005 mol acid * (1 mol base/2 mol acid) = 0.00025 mol base needed.
4. 0.00025 mol base * 100 g/mol = 0.025 grams base = 25 milligrams base.

 

[California Bear]

Hi, sorry to bump up this thread again, but I have a question regarding the first problem.

Shouldn't the balanced equation be:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O ... rather than what the OP wrote?

Thank you in advance!

 

[quicheduty]

Hi, sorry to bump up this thread again, but I have a question regarding the first problem.

Shouldn't the balanced equation be:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O ... rather than what the OP wrote?

Thank you in advance!

You're right!