Question

Calculate the ph of a solution containing an amphetamine concentration of 250 mg/l with amphetamine (c9h13n) is a weak base with a pkb of 4.2.

Answer 1

Answer:

pH = 10.49

Explanation:

The reaction is :

C₉H₁₃N(aq)  + H₂O(l)  ⇄  C₉H₁₃NH⁺(aq) + OH⁻(aq)   (1)

The dissociation constant (Kb) of the above reaction is:

$K_{b} = \frac{[C_{9}H_{13}NH^{+}][OH^{-}]}{[C_{9}H_{13}N]}$

This Kb can be caculated using the pkb:

$p_{K_{b}} = -log(K_{b}) \rightarrow K_{b} = 10^{-p_{K_{b}}} = 10^{- 4.2}= 6.31 \cdot 10^{-5}$

To find the pH of the reaction (1) first we need to convert the concentration of the amphetamine in mg/L to mol/L as follows:

$[C_{9}H_{13}N] = \frac{250 mg}{L} \cdot \frac{1 g}{1000 mg} \cdot \frac{1 mol}{135.206 g} = 1.85 \cdot 10^{-3} mol/L$  

Now, from the reaction (1) we have:

                      C₉H₁₃N(aq)  +  H₂O(l)  ⇄  C₉H₁₃NH⁺(aq) + OH⁻(aq)   (1)

At equilibrium:  1.85x10⁻³- x                         x                    x

$K_{b} = \frac{[C_{9}H_{13}NH^{+}][OH^{-}]}{[C_{9}H_{13}N]}$

$6.31 \cdot 10^{-5} = \frac{x*x}{1.85 \cdot 10^{-3} - x}$

$x^{2} - 6.31 \cdot 10^{-5}*(1.85 \cdot 10^{-3} - x) = 0$   (2)

Solving equation (2) for x, we have:

x =  0.0003116 = [OH⁻] = [C₉H₁₃NH⁺]

Therefore, the pH of reaction (1) is:

$pOH = -log([OH^{-}]) = -log(0.0003116) = 3.51$

$pH = 14 - pOH = 14 - 3.51 = 10.49$

I hope it helps you!  

Answer 2

Answer:

The pH of a solution containing an amphetamine (with  pKb of 4.2) concentration of 250 mg/l is  10.5

Explanation:

To solve the question we note that the

The mass of amphetamine = 250 mg/l

The molar mass of amphetamine  = 135.2062 g/mol

Number of moles of amphetamine  = (250 mg)/(135.2062 g/mol) = 1.85×10⁻³ moles    

That is the initial concentration of amphetamine = 1.85×10⁻³ M

We have $pK_b = -log K_b$

therefore

$K_b = 10^{-pK_b} = 10^{-4.2} = 6.3 \times 10^-^5$

$K_b=\frac{[C_9H_{14}N][OH^-]}{[C_9H_{13}N]}\\= 6.3 \times 10^-^5$

Which gives 6.3×10⁻⁵  $= \frac{x^{2} }{0.00185-x}$

$we get \\(0.00185-x)*6.3*10^{-5} = x^{2}(0.00185-x)*6.3*10^{-5} - x^{2} = 0$

x =  0.0003116 = [OH⁻] = [C₉H₁₃NH⁺]

However x = [OH⁻] = 3.116×10⁻⁴

pOH = -log[OH⁻] = -log(3.116×10⁻⁴) =   3.506

pH = 14 - pOH

= 14 - 3.506

= 10.49

≈ 10.5