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.

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

therefore

Which gives 6.3×10⁻⁵  

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

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:

This Kb can be caculated using the pkb:

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:

Now, from the reaction (1) we have:

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

At equilibrium:  1.85x10⁻³- x                         x                    x

  (2)

Solving equation (2) for x, we have:

x =  0.0003116 = [OH⁻] = [C₉H₁₃NH⁺]

Therefore, the pH of reaction (1) is:

I hope it helps you!  

i dont know the answe