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CaHeK987 [17]
3 years ago
8

A solution of pyridinium bromide has a pH of 3.00. What is the concentration of the pyridinium cation at equilibrium, in units o

f molarity? Express molarity to two decimal places.
Chemistry
1 answer:
Artemon [7]3 years ago
4 0

Answer:

The concentration of the pyridinium cation at equilibrium is 1.00×10⁻³ M

Explanation:

In water we have

C₅H₅NHBr + H₂O ⇒ C₅H₅NH+ + Br−

Pyridinium Bromide (C₅H₅NHBr) Dissociates Completely  Into C₅H₅NH+ And Br− as such it is a strong Electrolyte.

Therefore the number of moles of positive ion produced per mole of C₅H₅NHBr  is one

pH = - log [H₃O⁺] Therefore 10^-pH = [H₃O⁺] = concentration of C₅H₅NHBr

= 10⁻³ = 0.001M = concentration of C₅H₅NHBr

The concentration of C₅H₅NHBr is = 1.00×10⁻³ M to two places of decimal

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.500 mol of br2 and .500 mol of cl2 are placed in a .500L flask and allowed to reach equilibrium. at equilibrium the flask was f
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Answer : The value of K_c for the given reaction is, 0.36

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

As we know that the concentrations of pure solids and liquids are constant that is they do not change. Thus, they are not included in the equilibrium expression.

The given equilibrium reaction is,

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The expression of K_c will be,

K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

First we have to calculate the concentration of Br_2,Cl_2\text{ and }BrCl.

\text{Concentration of }Br_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }Cl_2=\frac{Moles}{Volume}=\frac{0.500mol}{0.500L}=1M

\text{Concentration of }BrCl=\frac{Moles}{Volume}=\frac{0.300mol}{0.500L}=0.6M

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K_c=\frac{[BrCl]^2}{[Br_2][Cl_2]}

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Therefore, the value of K_c for the given reaction is, 0.36

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