Help please! I am super confused as to how to solve this problem. Here\'s the qu

Question

Help please! I am super confused as to how to solve this problem. Here's the question:

Aniline, C6H5NH2, is a weak base that is a precursor for the indigo dye that gives blue jeans their color. (TBD) Its conjugate acid is called anilinium, C6H5NH3 + , has a Ka of 2.4 × 10-5 , and can also be used to make dyes and printer ink. If you were to make a 0.750 M solution of aniline, C6H5NH2, what would the pH and pOH of the solution be? You may assume the auto-ionization of water is negligible. If you make any other assumptions in your calculations, make sure to show them and justify why you did so.

Explanation / Answer

Solution :-

reaction equation

C6H5NH2 + H2O ---------- > C6H5N3+ + OH-

Kb= Kw/ ka

Kb = 1*10^-14 / 2.4*10^-5 = 4.17*10^-10

Lets make the ICE table

C6H5NH2 + H2O ---------- > C6H5NH3+ + OH-

0.750 M                                         0               0

-x                                                   +x              +x

0.750-x                                           x             x

Kb = [C6H5NH3+][OH-] /[C6H5NH2]

4.17*10^-10 = [x][x]/[0.750-x]

since the kb is very small we can neglect the x from denominator.

4.17*10^-10 * 0.750 = x^2

3.13*10^-10 =x^2

Taking square root of both side we get

1.77*10^-5 = x =[OH-]

pOH = -log[OH-]

pOH- log[1.77*10^-5]

pOH =4.75

pH + pOH = 14

pH= 14 – pOH

pH= 14 – 4.75

pH = 9.25

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