Calculate the pH at $25^{\circ} C$ of a 0.36 M solution of ethylammonium bromide $\left( C _2 H _5 NH _3 Br \right)$. Note that ethylamine $\left( C _2 H _5 NH _2\right)$ is a weak base with a $p K_b$ of 3.19.
Round your answer to 1 decimal place.
$pH=$
Answer (2)
Ethylammonium bromide dissociates in water, forming ethylammonium ions that act as a weak acid.
Calculate K b from p K b and then K a using the relationship K a = K b K w .
Use an ICE table to find the equilibrium concentration of H 3 O + , approximating that the change in concentration of C 2 H 5 N H 3 + is negligible.
Calculate the pH using p H = − l o g [ H 3 O + ] and round to one decimal place: 5.6 .
Explanation
Problem Analysis We are given a 0.36 M solution of ethylammonium bromide ( C 2 H 5 N H 3 B r ) . Ethylamine ( C 2 H 5 N H 2 ) is a weak base with p K b = 3.19 . We need to calculate the pH of the solution at 2 5 ∘ C . Since ethylammonium bromide is a salt of a weak base and a strong acid, the ethylammonium ion ( C 2 H 5 N H 3 + ) will act as a weak acid in water.
Dissociation and Hydrolysis Ethylammonium bromide dissociates completely in water: C 2 H 5 N H 3 B r ( a q ) → C 2 H 5 N H 3 + ( a q ) + B r − ( a q ) The ethylammonium ion acts as a weak acid and undergoes hydrolysis: C 2 H 5 N H 3 + ( a q ) + H 2 O ( l ) ⇌ C 2 H 5 N H 2 ( a q ) + H 3 O + ( a q )
Calculating Ka First, we need to calculate the K a for the ethylammonium ion. We know that K a = K b K w , where K w = 1.0 × 1 0 − 14 . We can calculate K b from p K b :
K b = 1 0 − p K b = 1 0 − 3.19 ≈ 6.4565 × 1 0 − 4 Now we can calculate K a :
K a = 6.4565 × 1 0 − 4 1.0 × 1 0 − 14 ≈ 1.5488 × 1 0 − 11
ICE Table Now we set up an ICE table to determine the equilibrium concentration of H 3 O + . Let x be the concentration of H 3 O + at equilibrium.
C 2 H 5 N H 3 +
C 2 H 5 N H 2
H 3 O +
Initial
0.36
0
0
Change
-x
+x
+x
Equilibrium
0.36 - x
x
x
Then, K a = [ C 2 H 5 N H 3 + ] [ C 2 H 5 N H 2 ] [ H 3 O + ] = 0.36 − x x 2 . Since K a is small, we can assume that 0.36 − x ≈ 0.36 . Therefore, K a = 0.36 x 2 .
Calculating [H3O+] Solving for x: x = K a × 0.36 = 1.5488 × 1 0 − 11 × 0.36 ≈ 2.3613 × 1 0 − 6 This is the concentration of H 3 O + at equilibrium.
Calculating pH Now we calculate the pH using the formula p H = − l o g [ H 3 O + ] :
p H = − l o g ( 2.3613 × 1 0 − 6 ) ≈ 5.6268 Rounding to 1 decimal place, we get p H = 5.6 .
Final Answer Therefore, the pH of the 0.36 M solution of ethylammonium bromide at 2 5 ∘ C is 5.6.
Examples
Understanding pH is crucial in many real-world applications. For instance, in agriculture, knowing the pH of the soil helps farmers determine the best crops to grow and the right fertilizers to use. In medicine, maintaining the correct pH balance in the body is essential for various biological processes. Similarly, in environmental science, monitoring the pH of water sources helps assess pollution levels and ensure water quality. This problem demonstrates how to calculate the pH of a solution, a fundamental skill in chemistry with broad practical implications.
The pH of a 0.36 M solution of ethylammonium bromide is calculated to be approximately 5.6. This is found by determining the K a of the ethylammonium ion and using an ICE table to find the equilibrium concentration of hydronium ions. The final pH is rounded to one decimal place as 5.6.
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