Respuesta :
Answer: The [tex]K_b[/tex] for the weak base is [tex]3.5\times 10^{-3}[/tex]
Explanation:
[tex]A^-+H_2O\rightarrow HA+OH^-[/tex]
cM 0 0
[tex]c-c\alpha[/tex] [tex]c\alpha[/tex] [tex]c\alpha[/tex]
So dissociation constant will be:
[tex]K_b=\frac{[HA][OH^-]}{[A^-]}[/tex]
[tex]K_b=\frac{(c\alpha)^{2}}{c-c\alpha}[/tex]
Given :
c= 0.0910 M and pH = 10.50
pOH = 14-pH = 14-10.50 = 3.5
Also [tex]pOH=-log[OH^-][/tex]
[tex][OH^-]=antilog(-3.5)= 3.2\times 10^{-4}M[/tex]
[tex][OH^-]=c\alpha=3.2\times 10^{-4}[/tex]
[tex]K_b=\frac{(3.2\times 10^{-4})^2}{(0.0910-3.2\times 10^{-4}}[/tex]
[tex]K_b=\frac{(3.2\times 10^{-4})^2}{(0.09068}[/tex]
[tex]K_b=3.5\times 10^{-3}[/tex]
Thus [tex]K_b[/tex] for the weak base is [tex]3.5\times 10^{-3}[/tex]
The Kb for the unknown weak base is 3.5 × 10⁻³.
How we calculate the dissociation constant for a weak base?
Dissociation constant for a weak base at equilibrium is the ratio of the product of concentration of product to the concentration of un-ionized base.
Chemical reaction for a weak base at initial and equilibrium state can be represented as follow:
[tex]\[\mathop {{{\rm{A}}^{\rm{ - }}}}\limits_{\scriptstyle{\rm{c}}\hfill\atop\scriptstyle{\rm{c - c\alpha }}\hfill} {\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\mathop { \to {\rm{HA}}}\limits_{\scriptstyle{\rm{0}}\hfill\atop\scriptstyle{\rm{c\alpha }}\hfill} {\rm{ + }}\mathop {{\rm{O}}{{\rm{H}}^{\rm{ - }}}}\limits_{\scriptstyle{\rm{0}}\hfill\atop\scriptstyle{\rm{c\alpha }}\hfill} \][/tex]
Kb = [HA] [OH⁻] / [A⁻]
Kb = (c∝)² / (c - c∝)
In the question given that,
pH = 10.50
Concentration of weak base c = 0.0910 M
We know that, pOH = 14 - pH
pOH = 14 - 10.50 = 3.5
Also we now that,
pOH = -log[OH⁻]
[OH⁻] = c∝ = antilog (-3.5) = 3.2 × 10⁻⁴ M
Now we put all these values in the above equation of Kb, we get
Kb = (3.2 × 10⁻⁴)² / (0.0910 - 3.2 × 10⁻⁴) = 3.5 × 10⁻³
Hence, 3.5 × 10⁻³ is the value of Kb.
To know more about dissociation constant, visit the below link:
https://brainly.com/question/3006391
