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

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

Step-by-step 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* 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* 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* 10^(-9)* 0.345=6.135*10^(-10)\\ \\ x=\sqrt{6.135*10^(-10)}=2.477* 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