Respuesta :
Answer:
The pH of the solution lies from 11 to 12.Hence, option e is correct.
Explanation:
The value of [tex]K_b[/tex] for caffine = [tex]4\times 10^{-4}[/tex]
[tex]CafOH(aq)\rightleftharpoons Caf(aq)+OH^-(aq)[/tex]
Initial
0 0.01 M 0
AT equilibrium:
x (0.01 -x)M x
[tex]K_b=\frac{x(0.01-x)}{(x)}[/tex]
[tex]4\times 10^{-4}=\frac{x(0.01-x)}{(x)}[/tex]
Solving for x:
x = 0.0096 M
The pOH of the solution is given by :
[tex]pOH=-\log[OH^-}[/tex]
[tex]pOH=-\log[x][/tex]
[tex]pOH=-\log[0.0096][/tex]
pOH = 2.02
pH= 14 - pOH = 14 - 2.02 = 11.98
The pH of the solution lies from 11 to 12.
pH is the measurement of the acidity and alkaline level of a solution. It indicates the levels of hydrogen ions present in the solution.
pH can be calculated as :
pH = - log [H₃O+]
The correct answer is :
Option D. 11-12
To calculate the pH of a solution we need to know the concentration of hydronium ion in moles per litre.
Given,
The value of Kb of caffeine= 4 × 10⁻⁴
[tex]\text {Caf OH}\; \text{(aq)} \rightleftharpoons \text{Caf (aq)} & \; + \text{OH}^{-} \text{(aq)}[/tex]
Initially:
0 0.01 M 0
At the equilibrium:
Y (0.01- Y)M Y
[tex]\text{K}_{\text{b}} & = \frac{\text{Y}(0.01 - \text{Y})}{\text{Y}}[/tex]
[tex]4 \times 10^{-4} = \frac{\text{Y}(0.01 - \text{Y})}{\text{Y}}[/tex]
Solving for Y:
Y = 0.0096 M
The pOH of the solution can be determined by:
[tex]\text{pOH} & = \text{- log}\; (\text{OH}^{-} )[/tex]
[tex]\text{pOH} & = \text{- log}\; (\text{Y} )[/tex]
[tex]\text{pOH} & = \text{- log}\; (0.0096)[/tex]
pOH = 2.02
pH = 14- pOH
= 14-2.02
= 11.98
Therefore, the pH of the solution ranges between 11 - 12.
To learn more on pH, acids and bases follow the link:
https://brainly.com/question/22228645
