Question

In each case write the chemical equation which describes the relevant acid-base reaction before you start the mathematics. What are the exact pHs of: A. 10^-2 M acetic acid B. 5 x 10 -2 M sodium acetate C. 10^-3 M pyridinium chloride Please show the steps so I can understand how to do it.

Answer 1

What are the exact pHs of:

A. 10^-2 M acetic acid   Ka = 1.8*10^-5

Solution :-

CH3COOH (aq) + H2O(l) ------- > CH3COO-(aq) + H3O+(aq)

10E-2                                                    0                         0

-x                                                             +x                      +x

10^-2-x                                                   x                          x

Ka = [H3O+][CH3COO-] / [CH3COOH]

1.8*10^-5 = [x][x]/[1*10^-2 –x]

Ka is very small therefore we can neglect the x from denominator then we get

1.8*10^-5 = [x][x]/[10^-2 ]

1.8*10^-5 * 10^-2 = x^2

1.8*10^-7 = x^2

Taking square root of both sides we get

4.24*10^-4 = x = [H3O+]

pH= -log [H3O+]

pH = -log [4.24*10^-4]

pH= 3.37

B. 5 x 10 -2 M sodium acetate ka = 5.55*10^-10

Solution :-

CH3COO- + H2O ------- > CH3COOH   + OH-

5*10^-2                                     0                  0

-x                                                +x                +x

5*10^-2 –x                                x                  x

Kb = [CH3COOH][OH-]/[CH3COO-]

5.55*10^-10 = [x][x]/[5*10^-2 –x]

Since kb is small so we can neglect the x from denominator then we get

5.55*10^-10 = [x][x]/[5*10^-2 ]

5.55*10^-10 * 5*10^-2 = x^2

2.77*10^-11 = x^2

Taking square root of both sides we get

5.26*10^-6 = x =[OH-]

pOH= -log [OH-]

pOH= -log[5.26*10^-6]

pOH= 5.28

pH + pOH = 14

pH= 14 – pOH

pH= 14 – 5.28

pH= 8.72

C. 10^-3 M pyridinium chloride ka = 5.88*10^-6

Solution :-

C5H6N + H2O ------- > C5H5N   + H2O

10^-3                                     0                  0

-x                                                +x                +x

10^-3 –x                                x                  x

Ka = [C5H6N][H3O+]/[C5H6N+]

5.88*10^-6 = [x][x]/[10^-3 –x]

Since kb is small so we can neglect the x from denominator then we get

5.88*10^-6 = [x][x]/[10^-3 ]

5.88*10^-6 * 10^-3 = x^2

5.88*10^-9 = x^2

Taking square root of both side we get

7.67*10^-5 = x= [H3O+]

pH= -log [H3O+]

pH= -log [7.67*10^-5]

pH= 4.12