Question

Determine the ph of a 0.227 m c5h5n solution at 25°c. the kb of c5h5n is 1.7 × 10-9.

Answer 1

The pH of the given 0.227 M solution of ${{\mathbf{C}}_{\mathbf{5}}}{{\mathbf{H}}_{\mathbf{5}}}{\mathbf{N}}$is$\boxed{{\text{9}}{\text{.3}}}$.

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:

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

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${{\text{K}}_{\text{b}}}$.

The molecule ${{\mathbf{C}}_{\mathbf{5}}}{{\mathbf{H}}_{\mathbf{5}}}{\mathbf{N}}$is a weak base that produces${{\mathbf{C}}_{\mathbf{5}}}{{\mathbf{H}}_{\mathbf{5}}}{\mathbf{N}}{{\mathbf{H}}^+}$ and hydroxide ion.

${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}\left({aq}\right)+{{\text{H}}_2}{\text{O}}\left(l\right)\rightleftharpoons{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}^+}\left({aq}\right)+{\text{O}}{{\text{H}}^-}\left({aq}\right)$

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

${{\text{K}}_{\text{b}}}=\frac{{\left[{{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}^+}}\right]\left[{{\text{O}}{{\text{H}}^-}}\right]}}{{\left[{{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}}\right]}}$                      …... (1)

The value of${{\text{K}}_{\text{b}}}$is ${\text{1}}{\text{.7}}\times{\text{1}}{{\text{0}}^{-9}}$.

The initial concentration of ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$is 0.227 M.

Let the change in concentration at equilibrium be x. Therefore, the concentration of ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}$becomes 0.227-x at equilibrium. The concentration of${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}^{\text{+}}}$and ${\text{O}}{{\text{H}}^-}$becomes x at equilibrium.

Substitute x for $\left[{{\text{O}}{{\text{H}}^-}}\right]$, x for $\left[{{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{H}}^{\text{+}}}}\right]$ and 0.227-x for $\left[{{{\text{C}}_{\text{5}}}{{\text{H}}_{\text{5}}}{\text{N}}}\right]$in equation (1).

${{\text{K}}_{\text{b}}}=\frac{{{\text{x}}\times{\text{x}}}}{{0.227-{\text{x}}}}$              …… (2)

Rearrange the equation (2) and substitute ${\text{1}}{\text{.7}}\times{\text{1}}{{\text{0}}^{-9}}$for ${{\text{K}}_{\text{b}}}$to calculate value of x.

${{\text{x}}^2}=\left({{\text{1}}{\text{.7}}\times{\text{1}}{{\text{0}}^{-9}}}\right)\left({{\text{0}}{\text{.227}}-{\text{x}}}\right)$

The final quadric equation is,

${{\text{x}}^2}-\left({{\text{1}}{\text{.7}}\times{\text{1}}{{\text{0}}^{-9}}}\right){\text{x}}-3.859\times{10^{-10}}=0$

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

${\text{x}}={\text{1}}{\text{.9645}}\times{\text{1}}{{\text{0}}^{-5}}$

The concentration of$\left[{{\text{O}}{{\text{H}}^-}}\right]$is equal to x and therefore the concentration of $\left[{{\text{O}}{{\text{H}}^-}}\right]$is ${\text{1}}{\text{.9645}}\times{\text{1}}{{\text{0}}^{-5}}{\text{ M}}$.

The formula to calculate the pH of the solution is,

${\text{pH}}=14-{\text{pOH}}$

This expression can be elaborate as,

$\begin{aligned}{\text{pH}}&=14-{\text{pOH}}\hfill\\{\text{pH}}&=14-\left({-\log\left[{{\text{O}}{{\text{H}}^-}}\right]}\right)\hfill\\{\text{pH}}&=14+\log\left[{{\text{O}}{{\text{H}}^-}}\right]\hfill\\\end{aligned}$      …… (3)

Substitute ${\text{1}}{\text{.9645}}\times{\text{1}}{{\text{0}}^{-5}}{\text{ M}}$for $\left[{{\text{O}}{{\text{H}}^-}}\right]$in equation (3)

$\begin{aligned}{\text{pH}}&=14+\log\left[{{\text{O}}{{\text{H}}^-}}\right]\\&=14+\log\left({{\text{1}}{\text{.9645}}\times{\text{1}}{{\text{0}}^{-5}}}\right)\\&={\mathbf{9}}{\mathbf{.3}}\\\end{aligned}$

Therefore, the pH of the given solution is 9.3.

Learn more:

1. The difference between heat and temperature.: brainly.com/question/914750

2. Complete equation for the dissociation of ${\text{N}}{{\text{a}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$(aq): brainly.com/question/5425813

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Equilibrium

Keywords: pH, 0.227 M, solution, c5h5N, c5h5N+, 9.3, pOH, equilibrium, pH, c5h5n and0.227 M-x.

Answer 2

 the concentration of OH- using the formula:
[OH-] = Kw / [H+]

[OH-] = 1.0 x 10^-14 / ( [(1.7 x 10^-9) (0.227)]^2 )

[OH-] = 5.1 x 10^-10

To find the pOH, use the formula:
pOH = -log [OH-]

pOH = -log [5.1 x 10^-10]
pOH = 9.29