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Sunny_sXe [5.5K]
3 years ago
7

Lithium diisopropylamide (LDA) is used as a strong base in organic synthesis. LDA is itself prepared by an acid-base reaction be

tween butyllithium and N,N-diisopropylamine, [(CH3)2CH]2NH. Write an equation for this reaction, and place the appropriate labels on each reagent and product from the following list: stronger acid, stronger base, weaker acid, weaker base.
Chemistry
1 answer:
kakasveta [241]3 years ago
5 0

Explanation:

Lithium diisopropylamide (LDA) is used in many organic synthesis and is a strong base. It is prepared by the acid base reaction of N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) and butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ).

The equation is show below as:

[(CH₃)₂CH]₂NH + Li⁺⁻CH₂CH₂CH₂CH₃  ⇒  [(CH₃)₂CH]₂N⁻Li⁺  + CH₃CH₂CH₂CH₃

N,N-diisopropylamine ( [(CH₃)₂CH]₂NH ) is a weaker acid and hence,  LDA ( [(CH₃)₂CH]₂N⁻Li⁺ ) is stronger base. (Weaker acid has stronger conjugate base)

Butyllithium ( Li⁺⁻CH₂CH₂CH₂CH₃ ) is a very strong base and hence, butane ( CH₃CH₂CH₂CH₃ ) is a very weak acid. (Strong base has weaker conjugate acid)

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Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral
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Answer:

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b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

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Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = \frac{0.350 g}{78 g/mol}=0.004487 mol

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Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

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The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

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Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

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Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

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Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

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1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

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