Calculate the pH at [tex]$25^{\circ} C$[/tex] of a 0.58 M solution of pyridinlum chloride [tex]$\left( C _5 H _5 NHCl \right)$[/tex]. Note that pyridine [tex]$\left( C _5 H _5 N\right)$[/tex] is a weak base with a [tex]$pK_b$[/tex] of 8.77. Round your answer to 1 decimal place.
Answer (1)
Determine that pyridinium chloride is the salt of a weak base and strong acid, so it will form a weak acid in solution.
Calculate p K a using the formula p K a = 14 − p K b , which gives p K a = 5.23 .
Calculate K a using K a = 1 0 − p K a , and set up an ICE table to find the concentration of H 3 O + .
Calculate the pH using the formula p H = − l o g [ H 3 O + ] , which results in p H = 2.7 .
Explanation
Problem Analysis We are given a 0.58 M solution of pyridinium chloride ( C 5 H 5 N H Cl ) , which is the salt of a weak base (pyridine, C 5 H 5 N ) and a strong acid (HCl). We are also given that the p K b of pyridine is 8.77. We need to find the pH of the solution.
Finding Ka Pyridinium chloride dissociates in water to give pyridinium ions ( C 5 H 5 N H + ) and chloride ions ( C l − ) . The pyridinium ion acts as a weak acid and hydrolyzes in water to produce pyridine and hydronium ions ( H 3 O + ). The chloride ion is a spectator ion and does not affect the pH. The hydrolysis reaction is: C 5 H 5 N H + ( a q ) + H 2 O ( l ) ⇌ C 5 H 5 N ( a q ) + H 3 O + ( a q ) We need to find the K a for the pyridinium ion. We know that p K a + p K b = 14 . Therefore, p K a = 14 − p K b = 14 − 8.77 = 5.23 Now we can calculate K a :
K a = 1 0 − p K a = 1 0 − 5.23
ICE Table and Solving for x Now, we set up an ICE table for the hydrolysis reaction: \t\t C 5 H 5 N H + ( a q ) + H 2 O ( l ) ⇌ C 5 H 5 N ( a q ) + H 3 O + ( a q ) Initial:\t0.58\t\t\t0\t\t\t0 Change:\t-x\t\t\t+x\t\t\t+x Equilibrium:\t0.58-x\t\t\tx\t\t\tx The expression for K a is: K a = [ C 5 H 5 N H + ] [ C 5 H 5 N ] [ H 3 O + ] = 0.58 − x x 2 Since K a is small, we can assume that 0.58 − x ≈ 0.58 . Then, K a = 0.58 x 2 Solving for x: x = K a × 0.58 = 1 0 − 5.23 × 0.58
Calculating pH Now we calculate the pH: p H = − l o g [ H 3 O + ] = − l o g ( x ) = − l o g ( 1 0 − 5.23 × 0.58 ) p H = − l o g ( 5.9 × 1 0 − 6 ) = 2.7
Final Answer The pH of the solution is 2.7.
Examples
Understanding pH is crucial in many real-world applications. For example, in agriculture, knowing the pH of the soil helps farmers choose the right crops and optimize growing conditions. In medicine, maintaining the correct pH balance in the body is essential for various biological processes. In environmental science, monitoring the pH of water sources helps assess pollution levels and protect aquatic life. In the food industry, pH control is vital for preserving food and ensuring its safety.
