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Olegator [25]
3 years ago
8

Can a substance be a lewis acid without being a bronsted-lowry acid?argue

Chemistry
2 answers:
storchak [24]3 years ago
6 0

Answer:

Yes

Explanation:

Yes, A substance can be a lewis acid without being a Bronsted-Lowery acid because there are some substances which cannot donate protons(Bronsted-Lowery acid) but can accept a pair of electron.

<u><em>For Example:</em></u>

Let us take the example of BF₃

BF₃ contains no proton so it is not a Bronsted Lowery Acid

However, BF₃ has an incomplete octet with 6 electrons. It needs an electron pair to complete its octet. It accepts a pair of electron to become a Lewis Acid

Zigmanuir [339]3 years ago
6 0

Answer:

Interesting question, and the answer is yes, a substance can be a Lewis acid but not a Bronsted-Lowrey acid. To see this, let’s take a look at the definitions of each.

Explanation:

Bronsted-Lowrey acid:

A compound that is a hydrogen ion (proton) donor. When dissolved in the solvent in question, these compounds lose a proton to the solution. The concentration of these protons in solution is referred to as acidity, and is measured on the pH scale.

Lewis acid:

A substance that is an electron pair receiver. In solution, free electron pairs will form bonds with the substance, either ionic or covalent. In this definition, a proton is itself an acid, rather than a part of an acid.

A key thing to note here is that, in the Bronsted-Lowrey definition, there must be a proton. That means, all Bronsted-Lowrey acids are of the form  HXn→H++Xn− , showing the dissociation in solution. However, a Lewis acid needs only to have the ability to accept an electron pair, which means that  H+  is a Lewis acid, instead of what makes a compound an acid. Additionally, that means that the number of compounds that qualify as a Lewis acid are expanded. A favorite example of mine is boron trifluoride, or  BF3 . It is a common reagent in organic synthesis, it is a Lewis acid, but does not have any hydrogen, so it cannot be a Bronsted-Lowrey acid

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0.84 moles

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3 years ago
You have 500.0 ml of a buffer solution containing 0.30 m acetic acid (ch3cooh) and 0.20 m sodium acetate (ch3coona). what will t
Nataly [62]
First, we should get moles acetic acid = molarity * volume

                                                                =0.3 M * 0.5 L

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then, we should get moles acetate = molarity * volume

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then, we have to get moles of OH- which added:

moles OH- = molarity  * volume

                   = 1 M    * 0.02L

                  = 0.02 mol

when the reaction equation is:


                 CH3COOH  +  OH-  → CH3COO-   +  H2O


moles acetic acid after adding OH- = (0.15-0.02) 
                                             
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moles acetate after adding OH- =  (0.1 + 0.02)

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Total volume = 0.5 L + 0.02 L= 0.52 L

∴[acetic acid] = moles acetic acid after adding OH- / total volume

                        = 0.13mol / 0.52L

                       = 0.25 M

and [acetate ) = 0.12 mol / 0.52L
 
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by using H-H equation we can get PH:

PH = Pka + ㏒[salt/acid]

when we have Ka = 1.8 x 10^-5

∴Pka = -㏒Ka 

        = -㏒ 1.8 x 10^-5

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So by substitution:

∴ PH = 4.7 + ㏒[acetate/acetic acid]

         = 4.7 + ㏒(0.23/0.25)

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Masteriza [31]
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tester [92]

Answer:

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