Question

What is the $pH$ of an aqueous solution of $0.364M$ ethylamine (a weak base with the formula ${C_2}{H_5}N{H_2}$)?

Answer 1

Hint: We have to know, $pH$ meaning 'capability of hydrogen' or 'force of hydrogen' is a scale used to determine the acidity or basicity of a fluid arrangement. Acidic are estimated to have lower $pH$ esteems than essential or basic arrangements.

Complete step by step answer:
We have to know that, ethylamine is a powerless base, which implies that it doesn't ionize totally in fluid answer for structure ethyl ammonium cations, ${C_2}{H_5}NH_3^ + $, and hydroxide anions, $O{H^ - }$ . A balance will be set up between the unionized ethylamine atoms and the two particles that outcome from its ionization. Utilize this harmony response as a base for an $ICE$ table to discover the balance convergence of hydroxide anions,
${C_2}{H_5}N{H_{2(aq)}} + {H_2}{O_{(l)}} \rightleftharpoons {C_2}{H_5}NH_{3(aq)}^ + + OH_{(aq)}^ - $
          \[\begin{array}{*{20}{c}}
  I&{\;\;\;\;\;\;\;\;0.364}&{\;\;\;\;\;\;\;\;\;\;0}&{\;\;\;\;\;\;\;\;\;\;\;\;\;\;0}
\end{array}\]
        \[\begin{array}{*{20}{c}}
  C&{\;\;\;\;\;\;\left( { - x} \right)}&{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( { + x} \right)}&{\;\;\;\;\;\;\;\;\;\;\left( { + x} \right)}
\end{array}\]
        \[\begin{array}{*{20}{c}}
  E&{\;\;\;\;\;0.364 - x}&{\;\;\;\;\;\;\;\;\;\;\;\;\;x}&{\;\;\;\;\;\;\;\;x}
\end{array}\]
Then, the base dissociation constant for the equilibrium,
${K_b} = \dfrac{{[{C_2}{H_5}NH_3^ + ][O{H^ - }]}}{{[{C_2}{H_5}N{H_2}]}}$
This above expression is equal to the,
$5.6 \times {10^{ - 4}} = \dfrac{{x.x}}{{0.364 - x}} = \dfrac{{{x^2}}}{{0.364 - x}}$
The above expression is rearranged to a quadratic equation,
${x^2} + (5.6 \times {10^{ - 4}})x - 2.0384 \times {10^{ - 4}} = 0$
Now we get two values, where one value is negative and another one is positive. The negative value is neglected.
$x = 0.014$
Here, $x$ is nothing but $[O{H^ - }]$
Therefore,
$[O{H^ - }] = 0.014M$
Here, we calculate the value of $pOH$ by using the following formula,
$pOH = - \log ([O{H^ - }])$
Then, applying the $[O{H^ - }]$ value in the above equation,
$pOH = - \log (0.014) = 1.85$
Finally, we have to calculate the $pH$ of a aqueous solution,
$pH + pOH = 14$
Applying $pOH$value, to get the value of $pH$,
$pH = 14 - 1.85 = 12.15$
Hence,
$pH = 12.15$

Note: We have to know that the $pH$ scale is logarithmic and contrarily shows the convergence of hydrogen particles in the arrangement. This is on the grounds that the equation used to ascertain $pH$ approximates the negative of the base $10$ logarithms of the molar concentration of hydrogen particles in the arrangement. All the more decisively, $pH$ is the negative of the base $10$ logarithms of the action of the hydrogen particle.