Question

Pyridine is a weak base that is used in the manufacture of pesticides and plastic resins. It is also a component of cigarette smoke. Pyridine ionizes in water as follows: C5H5N+H2O⇌C5H5NH++OH− The pKb of pyridine is 8.75. What is the pH of a 0.345 M solution of pyridine? Express the pH numerically to two decimal places.

Answer 1

Answer:

       $\large\boxed{\large\boxed{pH=9.39}}$

Explanation:

1. Chemical equation (given)

          $C_5H_5N+H_2O\rightleftharpoons C_5H_5NH^++OH^-$

2. Chemical equation with the phases:

         $C_5H_5N_{(aq)}+H_2O_{(l)}\rightleftharpoons C_5H_5NH^+_{(aq)}+OH^-_{(aq)}$

3. Equilibrium constant

Only the species in aqueous phase appear in the equilibrium constant.

       $K_{eq}=K_b=[C_5H_5NH][OH^-]/[C_5H_5N]$

4. Calculate Kb

                     $pK_b=-log(K_b)\\ \\ k_b=10^{-pK_b}\\ \\ K_b=10^{-8.75}\\ \\ K_b=1.778\times 10^{-9}$

5. Write the ICE (initial, change, equilibrium) table for the aqueous species

ICE table:

                            C₅H₅N       C₅H₅NH⁺       +OH⁻

Initial                    0.345M            0                 0

Change                   - x                 + x              + x

Equilibrium            0345 - x          x                 x

5. Substitute in the equilibrium constant

         $1.778\times 10^{-9}=x^2/(0.345-x)$

To solve you can neglect x in 0.345 - x, because x is much (very much) less than 0.345M.

             $x^2=1.778\times 10^{-9}\times 0.345=6.135\times10^{-10}\\ \\ x=\sqrt{6.135\times10^{-10}}=2.477\times 10^{-5}$

6. Calculate pOH

• pOH = - log [OH⁻] = - log (x) = -log (2.477 × 10⁻⁵) = 4.61

7. Calculate pH

• pH + pOH = 14

• pH = 14 - pOH = 14 - 4.61 = 9.39