Problem: (b) What is the pH of 0.88 M pyridinium chloride, C 5H5NHCl?

FREE Expert Solution

C₅H₅NHCl → C₅H₅NH⁺ + Cl^-

C₅H₅NH⁺ → positively charged amine → weak acid

C₅H₅NH⁺ + H₂O → C₅H₅NH + H₃O⁺

K_b for C₅H₅NH = 1.7x10^-9

$K_{w} = K_{a} \cdot K_{b}$

Assuming the reaction is at 25°C: K_w = 1.0x10^-14

Calculate K_b:

$\frac{K_{w}}{K_{b}} = \frac{K_{a} \cdot K_{b}}{K_{b}} K_{b} = \frac{K_{w}}{K_{a}} K_{b} = \frac{1.0 \times 10^{- 14}}{1.7 \times 10^{- 9}}$

K_b = 5.88x10^-6

K_b is also an equilibrium expression and can be calculated as:

$K_{b} = \frac{products}{reactants}$

Solids and liquids are ignored in equilibrium expressions

$K_{b} = \frac{[C_{5} H_{5} N] [H_{3} O^{+}]}{[C_{5} H_{5} {NH}^{+}]}$

$5.88 x 10^{- 6} = \frac{[x] [x]}{[0.88 - x]}$

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Problem Details

(b) What is the pH of 0.88 M pyridinium chloride, C ₅H₅NHCl?

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Problem: (b) What is the pH of 0.88 M pyridinium chloride, C 5H5NHCl?

FREE Expert Solution

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