Part A Pridine is a weak base that is used in the manufacture of pesticides and

Question

Part A Pridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Prndine ionizes in water as follows: C5H5N H2 C5H5NH OH The pKb of pyridine is 8.75. What is the pH of a 0.195 Msolution of pyridine? (Assume that the temperature is 25 C) Express the pH numerically to two decimal places. pH- 9.98 Submit My Answers Give Up Review part Hints Incorrect, one attempt remaining: Try Again You may have used the hydroxide ion concentration of neutral water

Explanation / Answer

A

This is a base in water so, let the base be "B" and HB+ the protonated base "HB+"

there are free OH- ions so, expect a basic pH

B + H2O <-> HB+ + OH-

The equilibirum Kb:

Kb = [HB+][OH-]/[B]

let "x" be OH- in solution

in equilibrium due to sotichiometry:

[HB+]= x= [OH-]

Acocunt for the dissolved base in solution vs. not in solution:

[B] = 0.195-x

Substitute in Kb

Kb = [HB+][OH-]/[B]

Kb = 10^-pKb = 10^-8.75 =1.7782*10^-9

1.7782*10^-9 = x*x/(0.195-x)

Solve for x, using quadratic equation

x = OH- = 1.857*10^-5

pOH = -log(OH-) = -log(  1.857*10^-5) = 4.7311

pH = 14-pOH = 14-4.7311

pH = 9.27

B)

similaly:

By definition, in equilibrium:

Ka = [H+][A-]/[HA]

Then

Assume that [H+] = [A-] = x … due to stoichiometry (i.e. 1 mol of H+ per each mol of A-)

Then, in equilibrium [HA] = M – x (i.e. the original concentration minus the acid in equilibrium)

Substitute

Ka = (x*x)/(M-x)

Ka = 10^-pKa = 10^-4.2 = 6.31*10^-5

6.31*10^-5 = x*x/(0.66-x)

solve for x (quadratic equation)

x = [H+] = 0.00642 M

[H+] = [A-] = x = 0.00642

The ionization:

% ionization = [H+] / M * 100% = 0.00642 /(0.66) * 100% = 0.9696 %

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