The pH of an aqueous solution of 0.189 M ammonium bromide, NH4Br (aq), is … This solution is… acidic, basic, neutral.
The Correct Answer and Explanation is:
Correct Answer:
The pH of a 0.189 M solution of ammonium bromide (NH₄Br) is acidic.
Explanation:
Ammonium bromide (NH₄Br) is a salt that dissociates completely in water into NH₄⁺ (ammonium ion) and Br⁻ (bromide ion):NH4Br (aq)→NH4+(aq)+Br−(aq)\text{NH}_4\text{Br (aq)} \rightarrow \text{NH}_4^+ (aq) + \text{Br}^- (aq)NH4Br (aq)→NH4+(aq)+Br−(aq)
To determine the pH of the solution, we must analyze how the ions behave in water:
- Br⁻ is the conjugate base of a strong acid (HBr), so it does not hydrolyze in water. It is pH neutral.
- NH₄⁺ is the conjugate acid of a weak base (NH₃), and it does hydrolyze in water. It donates a proton to water:
NH4++H2O⇌NH3+H3O+\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+NH4++H2O⇌NH3+H3O+
This reaction produces H₃O⁺ ions, which increase the acidity of the solution. As a result, the pH is less than 7.
To find the exact pH, we can calculate the concentration of H₃O⁺ ions using the acid dissociation constant Ka of NH₄⁺. The Kb of NH₃ (ammonia) is approximately 1.8 x 10⁻⁵, and we use the relationship: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}} \approx 5.56 \times 10^{-10}Ka=KbKw=1.8×10−51.0×10−14≈5.56×10−10
Using the ICE table and assuming x is small:Ka=x2[NH4+]⇒x2=Ka⋅0.189⇒x2=5.56×10−10⋅0.189K_a = \frac{x^2}{[NH_4^+]} \Rightarrow x^2 = K_a \cdot 0.189 \Rightarrow x^2 = 5.56 \times 10^{-10} \cdot 0.189Ka=[NH4+]x2⇒x2=Ka⋅0.189⇒x2=5.56×10−10⋅0.189×2≈1.05×10−10⇒x≈1.02×10−5 M=[H3O+]x^2 \approx 1.05 \times 10^{-10} \Rightarrow x \approx 1.02 \times 10^{-5} \text{ M} = [H_3O^+]x2≈1.05×10−10⇒x≈1.02×10−5 M=[H3O+]pH=−log(1.02×10−5)≈4.99\text{pH} = -\log(1.02 \times 10^{-5}) \approx 4.99pH=−log(1.02×10−5)≈4.99
Conclusion:
The pH is approximately 4.99, which is acidic. Therefore, this solution is acidic due to the hydrolysis of the ammonium ion.
