The pH of a 1.00 liter solution of the weak base nicotine ( K _b = 1.05×10^{- 6}) is 10.20.
How many moles of nicotine were dissolved to form this solution?

Question

The pH of a 1.00 liter solution of the weak base nicotine ( K _b = 1.05×10^{- 6}) is 10.20. How many moles of nicotine were dissolved to form this solution?

Quizplus · Accepted Answer

Using the equation for the dissociation of a weak base: 
NH2 + H2O ⇌ NH3+ + OH-
The equilibrium constant expression for this reaction is: 
Kb = ([NH3+][OH-])/[NH2]
Since Kb is known, we can solve for [NH2]. 
Kb = 1.05 × 10^-6 = ([NH3+][OH-])/[NH2]
We know that the pH is 10.20, which means that [OH-] = 10^-3.80. 
pH + pOH = 14 → 14 - 10.20 = 3.80 → [OH-] = 10^-3.80 
Substituting in the values: 
1.05 × 10^-6 = ([NH3+][10^-3.80])/[NH2]
Solving for [NH2]: 
[NH2] = [NH3+]/1.05 × 10^-6 × 10^3.80 
Since the solution contains 1.00 L of the weak base, we can also use the concentration of [NH2] to find the number of moles: 
Moles of NH2 = [NH2] × volume 
Moles of NH2 = [NH3+]/1.05 × 10^-6 × 10^3.80 × 1.00 L 
Moles of NH2 = 2.4 × 10^-2 mol 
Therefore, the answer is C.

The PH of a 1