In the reaction given below, water behaves as $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$

Question

Consider the equilibrium reactions:

$H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-$ $(K_1)$
$H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}$ $(K_2)$
$HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}$ $(K_3)$

The equilibrium constant, $K_c$, for the following dissociation:

$H_3PO_4 \rightleftharpoons 3H^+ + PO_4^{3-}$

is:

Answer 1

Step 1: Recall the Bronsted-Lowry idea.
A Bronsted-Lowry acid donates a proton ($H^+$); a base accepts one. We watch how each species changes.
Step 2: Write the reaction.
\[ NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^- \]
Step 3: Track the ammonia.
$NH_3$ becomes $NH_4^+$ by gaining a proton, so ammonia acts as the base.
Step 4: Track the water.
$H_2O$ becomes $OH^-$ by losing a proton. Losing $H^+$ is exactly what an acid does.
Step 5: Name water's role.
Because water donates a proton here, it behaves as a Bronsted-Lowry acid in this particular reaction.
Step 6: Choose the option.
So the correct description is a Bronsted-Lowry acid, option 2.
\[ \boxed{\text{A Bronsted-Lowry acid}} \]

In the reaction given below, water behaves as \(NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-\)

Show Hint

The Correct Option is B

Solution and Explanation

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