A 0.457-M aqueous solution of (CH3)2NH (dimethylamine) has a pH of 12.2. Calculate the pH of a buffer solution that is 0.457 M in (CH3)2NH and 0.280 M in (CH3)2NH2 .

pH = 10.95

Explanation:

To solve this question we must, as first, find pKb of dimethylamine. Then, using H-H equation we can solve the pH of the buffer:

pKb dimethylamine:

Based on the equilibrium:

(CH3)2NH(aq) + H2O(l) ⇄ (CH3)2NH2⁺(aq) + OH-(aq)

Kb is defined as:

Kb = [OH-] [(CH3)2NH2⁺] / [(CH3)2NH]

Both (CH3)2NH2⁺(aq) + OH- comes from the same equilibrium, that means:

[OH-] = [(CH3)2NH2⁺]

And: [(CH3)2NH] = 0.457M

[OH-] can be obtained from pH as follows:

14 -pH = pOH

14-12.2 = 1.8 =pOH

10^-pOH = [OH-] = 0.01585M

Replacing:

Kb = [0.01585M] [(0.01585M] / [0.457M]

Kb = 5.50x10⁻⁴

pKb = -logkb = 3.26

pH of the buffer:

Using H-H equation for bases:

pOH = pKb + log [conjugate acid] / [weak base]

pOH = 3.26 + log [0.280M] / [0.457M]

pOH = 3.05

pH = 14 - pOH

pH = 10.95

Wow that a lot of text for my tiny so i no wanna answer

Answerthe half-reaction having the greatest tendency to occur at the anode:     i₂(aq) + 2e⁻  ⇄ 2i⁻  ;   e = +0.53 v======oxidation occurs at the anode. we must therefore find the half-reaction with the greatest tendency to oxidize.the given values of e are reduction potentials (the form of the half-reactions are of reduction).the lower the reduction potential, the higher the potential to oxidize. the half-cell with the lowest reduction potential will have the greatest tendency to oxidize at the anode.the half-reaction having the greatest tendency must be    i₂(aq) + 2e⁻  ⇄ 2i⁻    e = +0.53 vbecause its reduction potential is the lowest out of the three.