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kolbaska11 [484]
3 years ago
6

A) For the following reaction, K > 1. Classify each of the reactants and products based on their strength as Bronsted-Lowry a

cids or bases.
C9H7N + HNO2Doublearrow.GIFC9H7NH+ + NO2-

a) HNO2 1) stronger acid

b) NO2- 2) weaker acid

c) C9H7NH+ 3) stronger base

d) C9H7N 4) weaker base

B)
For the following reaction, K < 1. Classify each of the reactants and products based on their strength as Bronsted-Lowry acids or bases.

C5H11NH+ + C6H5COO-Doublearrow.GIFC5H11N + C6H5COOH

a) C5H11NH+ 1) strongest acid

b) C6H5COO- 2) stongest base

c) C5H11N 3) weakest acid

d) C6H5COOH 4) weakest base
Chemistry
1 answer:
Aleks04 [339]3 years ago
4 0

Answer:

Explanation:

According to Bronsted-Lowry acids or base theory ,  the reagent capable of giving hydrogen ion or proton  will be acid and that which accepts hydrogen ion or proton  will be base .

C₉H₇N + HNO₂   ⇄    C₉H₇NH⁺ + NO₂⁻

If K > 1 , reaction is proceeding from left to right .

Hence HNO₂ is giving H⁺ or proton and C₉H₇N is accepting proton to form

C₉H₇NH⁺ .  

Hence HNO₂ is bronsted acid and C₉H₇N is bronsted base .

B )

when K < 1 , reaction above proceeds from right to left . That means

C₉H₇NH⁺ is giving H⁺ so it is a bronsted acid and NO₂⁻ is accepting H⁺ so it is a bronsted base .

Hence ,  NO₂⁻ is a bronsted base and C₉H₇NH⁺ is a bronsted acid .

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A chemist dissolves of pure hydroiodic acid in enough water to make up of solution. Calculate the pH of the solution. Be sure yo
Gre4nikov [31]

Answer:

1.76

Explanation:

There is some info missing. I think this is the original question.

<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Calculate the molarity of HI(aq)

M = mass of solute / molar mass of solute × liters of solution

M = 0.660 g / 127.91 g/mol × 0.300 L

M = 0.0172 M

Step 2: Write the acid dissociation reaction

HI(aq) ⇄ H⁺(aq) + I⁻(aq)

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Step 3: Calculate the pH

pH = -log [H⁺]

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