The pH of an aqueous solution of 0.226 M ammonium bromide, NH4Br (aq), is __. This solution is (ACIDIC/BASIC/NEUTRAL).
The Correct Answer and Explanation is:
Correct Answer:
The pH of a 0.226 M solution of ammonium bromide (NH₄Br) is approximately 4.82. This solution is acidic.
Explanation:
Ammonium bromide (NH₄Br) is a salt that forms when a weak base (ammonia, NH₃) reacts with a strong acid (hydrobromic acid, HBr). When this salt dissolves in water, it dissociates completely into its ions:NH₄Br (aq)→NH₄⁺ (aq)+Br⁻ (aq)\text{NH₄Br (aq)} \rightarrow \text{NH₄⁺ (aq)} + \text{Br⁻ (aq)}NH₄Br (aq)→NH₄⁺ (aq)+Br⁻ (aq)
- Br⁻ is the conjugate base of the strong acid HBr. It does not hydrolyze in water and has no effect on pH.
- NH₄⁺ is the conjugate acid of the weak base NH₃. It does hydrolyze in water, releasing H⁺ ions:
NH₄⁺ + H₂O⇌NH₃ + H₃O⁺\text{NH₄⁺ + H₂O} \rightleftharpoons \text{NH₃ + H₃O⁺}NH₄⁺ + H₂O⇌NH₃ + H₃O⁺
This reaction increases the concentration of hydrogen ions in the solution, making the solution acidic.
To calculate the pH, we use the Ka of ammonium ion. First, find Kb for NH₃:Kb (NH₃)=1.8×10−5\text{Kb (NH₃)} = 1.8 \times 10^{-5}Kb (NH₃)=1.8×10−5
Then use the relationship between Ka and Kb: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
Now use the Ka expression to solve for [H₃O⁺]:Ka=x20.226→x2=(5.56×10−10)(0.226)K_a = \frac{x^2}{0.226} \rightarrow x^2 = (5.56 \times 10^{-10})(0.226)Ka=0.226×2→x2=(5.56×10−10)(0.226)x=1.26×10−10≈1.12×10−5 Mx = \sqrt{1.26 \times 10^{-10}} \approx 1.12 \times 10^{-5} \text{ M}x=1.26×10−10≈1.12×10−5 M
Now calculate pH:pH=−log[H3O+]=−log(1.12×10−5)≈4.82\text{pH} = -\log[H₃O⁺] = -\log(1.12 \times 10^{-5}) \approx 4.82pH=−log[H3O+]=−log(1.12×10−5)≈4.82
Because the pH is less than 7, the solution is acidic.
