{"id":30935,"date":"2025-06-21T20:53:04","date_gmt":"2025-06-21T20:53:04","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=30935"},"modified":"2025-06-21T20:53:05","modified_gmt":"2025-06-21T20:53:05","slug":"the-ph-of-an-aqueous-solution-of-0-189-m-ammonium-bromide-nh4br-aq-is","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-ph-of-an-aqueous-solution-of-0-189-m-ammonium-bromide-nh4br-aq-is\/","title":{"rendered":"The pH of an aqueous solution of 0.189 M ammonium bromide, NH4Br (aq), is"},"content":{"rendered":"\n
The pH of an aqueous solution of 0.189 M ammonium bromide, NH4Br (aq), is \u2026 This solution is\u2026 acidic, basic, neutral.<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Correct Answer:<\/strong><\/h3>\n\n\n\n
The pH of a 0.189 M solution of ammonium bromide (NH\u2084Br) is acidic.<\/strong><\/p>\n\n\n\n\n\n\n\n
Explanation:<\/strong><\/h3>\n\n\n\n
Ammonium bromide (NH\u2084Br) is a salt<\/strong> that dissociates completely in water into NH\u2084\u207a (ammonium ion)<\/strong> and Br\u207b (bromide ion)<\/strong>:NH4Br (aq)\u2192NH4+(aq)+Br\u2212(aq)\\text{NH}_4\\text{Br (aq)} \\rightarrow \\text{NH}_4^+ (aq) + \\text{Br}^- (aq)NH4\u200bBr (aq)\u2192NH4+\u200b(aq)+Br\u2212(aq)<\/p>\n\n\n\n
To determine the pH of the solution, we must analyze how the ions behave in water:<\/p>\n\n\n\n
\n
Br\u207b<\/strong> is the conjugate base of a strong acid (HBr), so it does not<\/strong> hydrolyze in water. It is pH neutral<\/strong>.<\/li>\n\n\n\n
NH\u2084\u207a<\/strong> is the conjugate acid of a weak base (NH\u2083), and it does<\/strong> hydrolyze in water. It donates a proton to water:<\/li>\n<\/ul>\n\n\n\n
This reaction produces H\u2083O\u207a ions<\/strong>, which increase the acidity<\/strong> of the solution. As a result, the pH is less than 7<\/strong>.<\/p>\n\n\n\n
To find the exact pH, we can calculate the concentration of H\u2083O\u207a ions using the acid dissociation constant Ka<\/strong> of NH\u2084\u207a. The Kb<\/strong> of NH\u2083 (ammonia) is approximately 1.8 x 10\u207b\u2075<\/strong>, and we use the relationship:Ka=KwKb=1.0\u00d710\u2212141.8\u00d710\u22125\u22485.56\u00d710\u221210K_a = \\frac{K_w}{K_b} = \\frac{1.0 \\times 10^{-14}}{1.8 \\times 10^{-5}} \\approx 5.56 \\times 10^{-10}Ka\u200b=Kb\u200bKw\u200b\u200b=1.8\u00d710\u221251.0\u00d710\u221214\u200b\u22485.56\u00d710\u221210<\/p>\n\n\n\n
Using the ICE table and assuming x is small:Ka=x2[NH4+]\u21d2x2=Ka\u22c50.189\u21d2x2=5.56\u00d710\u221210\u22c50.189K_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\u200b=[NH4+\u200b]x2\u200b\u21d2x2=Ka\u200b\u22c50.189\u21d2x2=5.56\u00d710\u221210\u22c50.189×2\u22481.05\u00d710\u221210\u21d2x\u22481.02\u00d710\u22125 M=[H3O+]x^2 \\approx 1.05 \\times 10^{-10} \\Rightarrow x \\approx 1.02 \\times 10^{-5} \\text{ M} = [H_3O^+]x2\u22481.05\u00d710\u221210\u21d2x\u22481.02\u00d710\u22125 M=[H3\u200bO+]pH=\u2212log\u2061(1.02\u00d710\u22125)\u22484.99\\text{pH} = -\\log(1.02 \\times 10^{-5}) \\approx 4.99pH=\u2212log(1.02\u00d710\u22125)\u22484.99<\/p>\n\n\n\n
Conclusion:<\/strong><\/h3>\n\n\n\n
The pH is approximately 4.99<\/strong>, which is acidic<\/strong>. Therefore, this solution is acidic<\/strong> due to the hydrolysis of the ammonium ion.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
The pH of an aqueous solution of 0.189 M ammonium bromide, NH4Br (aq), is \u2026 This solution is\u2026 acidic, basic, neutral. The Correct Answer and Explanation is: Correct Answer: The pH of a 0.189 M solution of ammonium bromide (NH\u2084Br) is acidic. Explanation: Ammonium bromide (NH\u2084Br) is a salt that dissociates completely in water into […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-30935","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30935","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=30935"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30935\/revisions"}],"predecessor-version":[{"id":30943,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/30935\/revisions\/30943"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=30935"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=30935"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=30935"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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<\/figure>\n","protected":false},"excerpt":{"rendered":"