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Question: What is the \(pH\) of an aqueous solution of \(0.364M\) ethylamine (a weak base with the formula \({...

What is the pHpH of an aqueous solution of 0.364M0.364M ethylamine (a weak base with the formula C2H5NH2{C_2}{H_5}N{H_2})?

Explanation

Solution

We have to know, pHpH 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 pHpH 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, C2H5NH3+{C_2}{H_5}NH_3^ + , and hydroxide anions, OH−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 ICEICE table to discover the balance convergence of hydroxide anions,
C2H5NH2(aq)+H2O(l)⇌C2H5NH3(aq)++OH(aq)−{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,
Kb=[C2H5NH3+][OH−][C2H5NH2]{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,
5.6×10−4=x.x0.364−x=x20.364−x5.6 \times {10^{ - 4}} = \dfrac{{x.x}}{{0.364 - x}} = \dfrac{{{x^2}}}{{0.364 - x}}
The above expression is rearranged to a quadratic equation,
x2+(5.6×10−4)x−2.0384×10−4=0{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=0.014x = 0.014
Here, xx is nothing but [OH−][O{H^ - }]
Therefore,
[OH−]=0.014M[O{H^ - }] = 0.014M
Here, we calculate the value of pOHpOH by using the following formula,
pOH=−log⁡([OH−])pOH = - \log ([O{H^ - }])
Then, applying the [OH−][O{H^ - }] value in the above equation,
pOH=−log⁡(0.014)=1.85pOH = - \log (0.014) = 1.85
Finally, we have to calculate the pHpH of a aqueous solution,
pH+pOH=14pH + pOH = 14
Applying pOHpOHvalue, to get the value of pHpH,
pH=14−1.85=12.15pH = 14 - 1.85 = 12.15
Hence,
pH=12.15pH = 12.15

Note: We have to know that the pHpH 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 pHpH approximates the negative of the base 1010 logarithms of the molar concentration of hydrogen particles in the arrangement. All the more decisively, pHpH is the negative of the base 1010 logarithms of the action of the hydrogen particle.