Further problems on pH

 Hello dear! Welcome to today’s lesson. Do not hesitate to reach out to us via Whatsapp, Messenger or comments on the page if you need clarification on any of the problems that have been solved here. Today, we would continue as usual with mathematical examples on pH and we would be done with all these tomorrow. On Saturday, we shall begin to discuss about salts which would ultimately culminate in a discussion about  indicators as well as the concept of buffers. Stay tuned as we take this long ride!


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Example 6;

The pKb of pyridine is 8.77. What is the pH of a 0.010 M aqueous solution of  pyidine.


Solution;

We  must recall that pyridine is a weak base so the use of an ICE table is indispensable when we try obtain the equilibrium concentrations that would go into the equilibrium equation. We have to set up the ICE table as follows;


              C5H5N(aq) + H2O(l)    <-> C5H5NH^+(aq) + OH^- (aq)

I             0.010                                       0                           0

C            -x                                            +x                         +x       

E            0.010 – x                                 x                           x

Kb = [C5H5NH^+] [OH^-]/[C5H5N]

Again; 

Kb = Antilog (-pKb)

Kb = Antilog (-8.77)

Kb = 1.69 * 10^-9

Let [C5H5NH^+] = [OH^-] = x

1.69 * 10^-9 = x^2/ 0.010 – x

1.69 * 10^-9(0.010 – x) = x^2

1.69 * 10^-11 - 1.69 * 10^-9x = x^2

x^2 + 1.69 * 10^-9x - 1.69 * 10^-11= 0

x = 4.1 * 10^-6 M 

(Recall that I have always said that we leave the negative result of the quadratic behind since concentration can not be negative)

We now have that;

[C5H5NH^+] = [OH^-] = 4.1 * 10^-6 M 

pOH = -log(4.1 * 10^-6 M )

pOH = 5.4

pH = 14 – 5.4

pH = 8.6

Hence, the pH of the pyridine solution is 8.6.


Example 7;

Calculate the pH of a solution containing 545 mg/L of morphine (C17H19NO3). Morphine is a weak base with pKb 5.80.


Solution;

The unit given as mg/L is more common in environmental chemistry and it is mostly referred to as parts per million (ppm). We can convert from parts per million unit of concentration to the SI unit of mol/L(M) using the formula;

ppm * 0.001/molar mass 

Molar mass of morphine = 285 g/mol

Molar concentration of morphine = 545 * 0.001/285

=0.0019 M

Then we proceed to set up the ICE table or the ionization of morphine.

      C17H19NO3(aq) + H2O(l)  <-> C17H20NO3^+(aq) + OH^-(aq)

I       0.0019                                             0                               0

C        -x                                                 +x                            +x

E    0.0019 – x                                        x                               x

Kb = [C17H20NO3^+] [OH^-]/[C17H19NO3]

Let [C17H20NO3^+] = [OH^-] = x

From the question;

Kb = Antilog (-5.80)

Kb = 1.58 * 10^-6

Now;

1.58 * 10^-6 = x^2/(0.0019 – x)

1.58 * 10^-6(0.0019 – x) = x^2

3 * 10^-9 - 1.58 * 10^-6x = x^2

x^2 + 1.58 * 10^-6x - 3 * 10^-9= 0

x = 5.4 * 10^-5 M

Thus;

 [C17H20NO3^+] = [OH^-] =5.4 * 10^-5 M

pOH = -log (5.4 * 10^-5 M)

pOH = 4.3

pH = 14 – 4.3

pH = 9.7

The pH of the morphine solution is 9.7.


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