Respuesta :
Answer:
11.39
Explanation:
Given that:
[tex]pK_{b}=4.82[/tex]
[tex]K_{b}=10^{-4.82}=1.5136\times 10^{-5}[/tex]
Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{1.805\ g}{82.0343\ g/mol}[/tex]
[tex]Moles= 0.022\ moles[/tex]
Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
[tex]Molarity=\frac{0.022}{0.055}[/tex]
Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
[tex]K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}[/tex]
[tex]1.5136\times 10^{-5}=\frac {x^2}{0.4-x}[/tex]
x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
pH = 14 - pOH = 14 - 2.61 = 11.39
