Question

Caffeine (C_8H_10N_4O_2) is a weak base with a K_b value of 4 times 10^-4. The pH of a 0.01 M solution of caffeine is in the range of: a. 2-3 b. 5-6 c. 7-8 d. 9-10 e. 11-12

Answer 1

Answer:

The pH of the solution lies from 11 to 12.Hence, option e is correct.

Explanation:

The value of $K_b$ for caffine = $4\times 10^{-4}$

$CafOH(aq)\rightleftharpoons Caf(aq)+OH^-(aq)$

Initial

   0           0.01 M       0

AT equilibrium:

  x          (0.01 -x)M      x

$K_b=\frac{x(0.01-x)}{(x)}$

$4\times 10^{-4}=\frac{x(0.01-x)}{(x)}$

Solving for x:

x = 0.0096 M

The pOH of the solution is given by :

$pOH=-\log[OH^-}$

$pOH=-\log[x]$

$pOH=-\log[0.0096]$

pOH = 2.02

pH= 14 - pOH = 14 - 2.02 = 11.98

The pH of the solution lies from 11 to 12.

Answer 2

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

$\text {Caf OH}\; \text{(aq)} \rightleftharpoons \text{Caf (aq)} & \; + \text{OH}^{-} \text{(aq)}$

Initially:

        0             0.01 M            0

At the equilibrium:

       Y            (0.01- Y)M         Y

$\text{K}_{\text{b}} & = \frac{\text{Y}(0.01 - \text{Y})}{\text{Y}}$

$4 \times 10^{-4} = \frac{\text{Y}(0.01 - \text{Y})}{\text{Y}}$

Solving for Y:

Y = 0.0096 M

The pOH of the solution can be determined by:

$\text{pOH} & = \text{- log}\; (\text{OH}^{-} )$

$\text{pOH} & = \text{- log}\; (\text{Y} )$

$\text{pOH} & = \text{- log}\; (0.0096)$

pOH = 2.02

pH = 14- pOH

= 14-2.02

= 11.98

Therefore, the pH of the solution ranges between 11 - 12.

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