Question

Dimethylamine, (ch3)2nh, has a kb value of 5.4 x 10â4. in a 1.2 m solution of dimethylamine at 25Ëc what is the concentration of h3o+?

Answer 1

Answer is: 
concentration of H₃O⁺ is 4·10⁻¹³ mol/L.

Chemical reaction: (CH₃)₂NH + H₂O → (CH₃)₂NH₂⁺ +OH⁻.Kb((CH₃)₂NH) = 5,4·10⁻⁴.c((CH₃)₂NH) = 1,2 mol/L.Kb((CH₃)₂NH) = c(OH⁻) · c((CH₃)₂NH₂⁺) ÷ c((CH₃)₂NH).c(OH⁻) = c((CH₃)₂NH₂⁺) = x.5,4·10⁻⁴ = x² ÷ 1,2-x.

Solve quadratic equation:
x = 0,025 mol/L.c(OH⁻) · c(H₃O⁺) = 10⁻¹⁴.c(H₃O⁺) = 10⁻¹⁴ ÷ 0,025 = 4·10⁻¹³ mol/L.

Answer 2

The concentration of ${{\text{H}}_3}{{\text{O}}^ + } is$$\boxed{4 \times {{10}^{ - 13}}\;{\text{M}}}$

Further explanation:

Chemical equilibrium is the state in which concentration of reactants and products become constant and do not change with time. This is because the rate of forward and backward direction becomes equal. The general equilibrium reaction is as follows:

${\text{A(g)}}+{\text{B(g)}}\rightleftharpoons{\text{C(g)}}+{\text{D(g)}}$

Equilibrium constant is the constant that relates the concentration of product and reactant at equilibrium. The formula to calculate the equilibrium constant for general reaction is as follows:

${\text{K}}=\dfrac{{\left[ {\text{D}}\right]\left[{\text{C}}\right]}}{{\left[{\text{A}} \right]\left[{\text{B}}\right]}}$

Here, K is the equilibrium constant.

The equilibrium constant for the dissociation of acid is known as ${{\text{K}}_{\text{a}}}$ and equilibrium constant for the dissociation of base is known as ${{\text{K}}_{\text{b}}}$ .

For example, dimethylamine is a weak base that dissociates in water to form  ${\left({{\text{C}}{{\text{H}}_3}}\right)_2}{\text{N}}{{\text{H}}_2}^+$ and hydroxide ion.

${\left({{\text{C}}{{\text{H}}_3}}\right)_2}{\text{NH}}+{{\text{H}}_2}{\text{O}} \rightleftharpoons {\left( {{\text{C}}{{\text{H}}_3}}\right)_2}{\text{N}}{{\text{H}}_2}^ ++{\text{O}}{{\text{H}}^-}$

The expression of  ${{\text{K}}_{\text{b}}}$ for the above reaction is as follows:

${{\text{K}}_b}=\dfrac{{\left[{{{\left({{\text{C}}{{\text{H}}_3}}\right)}_2}{\text{N}}{{\text{H}}_2}^+} \right]\left[\\{{\text{O}}{{\text{H}}^-}}\right]}}{{\left[{{{\left({{\text{C}}{{\text{H}}_3}}\right)}_2}{\text{NH}}}\right]}}$       ...... (1)

The value of ${{\text{K}}_{\text{b}}}$ is ${\text{5}}{\text{.4}}\times{\text{1}}{{\text{0}}^{ - 4}}$ .

The initial concentration of dimethylamine is 1.2 M.

The change in the concentration at equilibrium is x. Therefore the concentration of dimethylamine becomes 1.2-x at equilibrium. The concentration of ${\left({{\text{C}}{{\text{H}}_3}}\right)_2}{\text{N}}{{\text{H}}_2}^+$ and ${\text{O}}{{\text{H}}^ - }$ becomes x at equilibrium.

Substitute x for $\left[ {{\text{O}}{{\text{H}}^ - }}\right]$ , x for  $\left[ {{{\left( {{\text{C}}{{\text{H}}_3}} \right)}_2}{\text{N}}{{\text{H}}_2}^ + } \right]$ and 1.2-x for $\left[{{{\left({{\text{C}}{{\text{H}}_3}}\right)}_2}{\text{NH}}} \right]$ in equation (1).

${{\text{K}}_{\text{b}}}=\dfrac{{{\text{x}}\times{\text{x}}}}{{1.2-{\text{x}}}}$        ...... (2)

Rearrange the equation (2) and substitute ${\text{5}}{\text{.4}} \times{\text{1}}{{\text{0}}^{ - 4}}$ for ${{\text{K}}_{\text{b}}}$ to calculate value of x.

$\begin{gathered}{{\text{x}}^2} =\left( {{\text{5}}{\text{.4}}\times {{10}^{ - 4}}} \right)\left( {{\text{1}}{\text{.2}} - {\text{x}}} \right)\\{{\text{x}}^2}-\left( {{\text{5}}{\text{.4}} \times {{10}^{ - 4}}}\right){\text{x}}-6.48\times{10^{ - 4}}\end{gathered}$

After solving the quadratic equation the value of x obtained is,

x = 0.02572

The concentration $\left[{{\text{O}}{{\text{H}}^ - }}\right]$ of is equal to x and therefore the concentration of $\left[{{\text{O}}{{\text{H}}^ - }}\right]$ is 0.02572 M.

The relation between the concentration of hydronium ion and hydroxide ion is as follows:

$\left[{{{\text{H}}_{\text{3}}}{{\text{O}}^ + }}\right]\left[{{\text{O}}{{\text{H}}^ - }}\right] = {10^{ - 14}}$                      ...... (3)

Here,

$\left[{{{\text{H}}_3}{{\text{O}}^ + }}\right]$ is the concentration of hydronium ion.

$\left[{{\text{O}}{{\text{H}}^ - }}\right]$ is the concentration of hydroxide ion.

Rearrange the equation (3) to calculate the concentration of hydronium ion.

$\left[ {{{\text{H}}_{\text{3}}}{{\text{O}}^ + }} \right]=\dfrac{{{{10}^{ - 14}}}}{{\left[ {\text{O}}{{\text{H}}^ - }}\right]}}$    …… (4)                                                                                              

Substitute 0.02572 for $\left[{{\text{O}}{{\text{H}}^ - }}\right]$ in the equation (4).

$\begin{aligned}\left[{{{\text{H}}_{\text{3}}}{{\text{O}}^ + }}\right]&=\frac{{{{10}^{ - 14}}}}{{0.02572}}\\&= 3.8880\times{10^{ - 13}}\;{\text{M}}\\&=4\times {10^{ - 13}}\;{\text{M}}\\\end{aligned}$

Learn more:

1. What is the equilibrium constant for pure water at ${\text{2}}{{\text{5}}^ \circ }{\text{C}}$ brainly.com/question/3467841

2. Determine the pH of 0.25 m methylamine.brainly.com/question/9040743

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Chemical equilibrium

Keywords: equilibrium, concentration, reactants, products, constant, ionization, dimethylamine, weak base, acid, 0.02572 M , hydroxide ion and hydronium ion.