Ammonium chloride (NH4Cl) is an acidic salt. When ammonium chloride is dissolved in water, it dissociates into its component ions. This reaction goes to completion, as indicated by the one-way arrow in the following equation: NH4Cl(s) ==========> NH4+(aq) + Cl-(aq) The Cl- ion, being the conjugate base of a strong acid (HCl), is too weak to react with water. However, the NH4+, being the conjugate acid of a weak base (NH3), is strong enough to react slightly with water. The equation for this equilibrium reaction is: NH4+(aq) + H2O(l) <———-> H3O+(aq) + NH3(aq) The equilibrium constant for the above reaction (Ka for NH4+) is normally not published in tables because it can be calculated from two other values that ARE published: the ionization constants for H2O and NH3. 2H2O(l) <———-> H3O+(aq) + OH-(aq) Kw = 1.0 x 10^-14 NH3(aq) + H2O(l) <———-> NH4+(aq) + OH-(aq) Kb = 1.8 x 10^-5 By combining these two equations in the appropriate way, it is possible to obtain the desired reaction, and hence, its equilibrium constant. What is the pOH of a 3.25 M NH4Cl solution? answers: 2.33 4.37 6.28 7.52 9.63 Can someone please explain how to do this step by step? Thank you.
The Correct Answer and Explanation is:
Step 1: Understand the Chemistry
Ammonium chloride (NH₄Cl) is a salt formed from a weak base (NH₃) and a strong acid (HCl). When it dissolves in water:NH₄Cl (s)→NH₄⁺ (aq)+Cl⁻ (aq)\text{NH₄Cl (s)} \rightarrow \text{NH₄⁺ (aq)} + \text{Cl⁻ (aq)}NH₄Cl (s)→NH₄⁺ (aq)+Cl⁻ (aq)
- The Cl⁻ ion is the conjugate base of a strong acid (HCl), so it does not hydrolyze.
- The NH₄⁺ ion is the conjugate acid of a weak base (NH₃), so it does hydrolyze and contributes to the acidity of the solution.
NH₄⁺ + H₂O↔H₃O⁺ + NH₃\text{NH₄⁺ + H₂O} \leftrightarrow \text{H₃O⁺ + NH₃}NH₄⁺ + H₂O↔H₃O⁺ + NH₃
Step 2: Determine KaK_aKa for NH₄⁺
We use the relationship:Ka×Kb=KwK_a \times K_b = K_wKa×Kb=Kw
Given:
- KbK_bKb for NH₃ = 1.8×10−51.8 \times 10^{-5}1.8×10−5
- KwK_wKw = 1.0×10−141.0 \times 10^{-14}1.0×10−14
Ka=KwKb=1.0×10−141.8×10−5=5.56×10−10K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}Ka=KbKw=1.8×10−51.0×10−14=5.56×10−10
Step 3: Set Up ICE Table
Initial concentration of NH₄⁺ = 3.25 M
Let xxx be the concentration of H₃O⁺ formed:NH₄⁺+H₂O↔H₃O⁺+NH₃Initial (M):3.2500Change (M):−x+x+xEquilibrium (M):3.25−xxx\begin{align*} \text{NH₄⁺} + \text{H₂O} &\leftrightarrow \text{H₃O⁺} + \text{NH₃} \\ \text{Initial (M)}: &\quad 3.25 \quad 0 \quad 0 \\ \text{Change (M)}: &\quad -x \quad +x \quad +x \\ \text{Equilibrium (M)}: &\quad 3.25 – x \quad x \quad x \\ \end{align*}NH₄⁺+H₂OInitial (M):Change (M):Equilibrium (M):↔H₃O⁺+NH₃3.2500−x+x+x3.25−xxx
Now apply KaK_aKa:Ka=x23.25−x≈x23.25K_a = \frac{x^2}{3.25 – x} \approx \frac{x^2}{3.25}Ka=3.25−xx2≈3.25×25.56×10−10=x23.25⇒x2=1.807×10−9⇒x=1.807×10−9=4.25×10−55.56 \times 10^{-10} = \frac{x^2}{3.25} \Rightarrow x^2 = 1.807 \times 10^{-9} \Rightarrow x = \sqrt{1.807 \times 10^{-9}} = 4.25 \times 10^{-5}5.56×10−10=3.25×2⇒x2=1.807×10−9⇒x=1.807×10−9=4.25×10−5
Step 4: Calculate pH and pOH
[H₃O⁺]=x=4.25×10−5⇒pH=−log(4.25×10−5)=4.37⇒pOH=14−4.37=9.63[\text{H₃O⁺}] = x = 4.25 \times 10^{-5} \Rightarrow \text{pH} = -\log(4.25 \times 10^{-5}) = 4.37 \Rightarrow \text{pOH} = 14 – 4.37 = 9.63[H₃O⁺]=x=4.25×10−5⇒pH=−log(4.25×10−5)=4.37⇒pOH=14−4.37=9.63
Final Answer: 9.63\boxed{9.63}9.63
Explanation
When ammonium chloride (NH₄Cl) dissolves in water, it fully dissociates into ammonium ions (NH₄⁺) and chloride ions (Cl⁻). The chloride ion, as the conjugate base of a strong acid (HCl), is essentially inert in water and does not affect the pH. The ammonium ion, however, is the conjugate acid of the weak base ammonia (NH₃), and thus it undergoes a slight hydrolysis with water to produce hydronium ions (H₃O⁺) and ammonia.
The acidity from NH₄⁺ hydrolysis is determined by its acid dissociation constant, KaK_aKa, which is not typically found in tables. However, since NH₄⁺ and NH₃ are conjugate acid-base pairs, KaK_aKa for NH₄⁺ can be calculated using the relationship Ka=KwKbK_a = \frac{K_w}{K_b}Ka=KbKw, where KbK_bKb is the base constant for ammonia and KwK_wKw is the ionization constant for water.
After calculating KaK_aKa, we set up an ICE table to determine the concentration of H₃O⁺ formed at equilibrium. This concentration allows us to find the pH, and subsequently the pOH using the relationship pH+pOH=14\text{pH} + \text{pOH} = 14pH+pOH=14. In a 3.25 M NH₄Cl solution, the final pOH comes out to 9.63, confirming that the solution is weakly acidic, as expected from a salt of a weak base and strong acid.
