{"id":189046,"date":"2025-02-08T11:17:27","date_gmt":"2025-02-08T11:17:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189046"},"modified":"2025-02-08T11:17:29","modified_gmt":"2025-02-08T11:17:29","slug":"classify-nh4no3-as-an-acidic-basic-or-neutral-salt-and-then-answer-the-following-question","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/08\/classify-nh4no3-as-an-acidic-basic-or-neutral-salt-and-then-answer-the-following-question\/","title":{"rendered":"Classify NH4NO3 as an acidic, basic, or neutral salt, and then answer the following question"},"content":{"rendered":"\n

Classify NH4NO3 as an acidic, basic, or neutral salt, and then answer the following question. What is the pH of 0.03 M NH,NO, (aq)?<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To classify NH\u2084NO\u2083 as an acidic, basic, or neutral salt, let’s break it down into its constituent ions: NH\u2084\u207a (ammonium ion) and NO\u2083\u207b (nitrate ion).<\/p>\n\n\n\n

    \n
  1. NH\u2084\u207a (Ammonium ion)<\/strong>: The ammonium ion is derived from the weak base ammonia (NH\u2083). When dissolved in water, ammonium ions undergo hydrolysis, meaning they react with water to form ammonia (NH\u2083) and hydronium ions (H\u2083O\u207a), which lowers the pH of the solution. This makes NH\u2084\u207a a weak acid. The reaction is as follows:
    [
    NH_4^+ + H_2O \\rightleftharpoons NH_3 + H_3O^+
    ]<\/li>\n\n\n\n
  2. NO\u2083\u207b (Nitrate ion)<\/strong>: The nitrate ion is the conjugate base of a strong acid, nitric acid (HNO\u2083). Conjugate bases of strong acids, such as NO\u2083\u207b, do not hydrolyze in water and have a neutral effect on pH.<\/li>\n<\/ol>\n\n\n\n

    Given that NH\u2084\u207a is acidic and NO\u2083\u207b is neutral, the salt NH\u2084NO\u2083 is acidic overall. This is because the contribution of the ammonium ion to lowering the pH outweighs any neutral effect of the nitrate ion.<\/p>\n\n\n\n

    pH of a 0.03 M NH\u2084NO\u2083 Solution<\/h3>\n\n\n\n

    To find the pH of a 0.03 M NH\u2084NO\u2083 solution, we need to consider the hydrolysis of the ammonium ion (NH\u2084\u207a). The relevant equilibrium for NH\u2084\u207a is:<\/p>\n\n\n\n

    [
    NH_4^+ + H_2O \\rightleftharpoons NH_3 + H_3O^+
    ]<\/p>\n\n\n\n

    The equilibrium constant for this reaction, ( K_b ), can be found using the relationship between the base dissociation constant of ammonia (NH\u2083) and the acid dissociation constant of NH\u2084\u207a. The value for ( K_b ) for ammonia is approximately ( 1.8 \\times 10^{-5} ), and the value for ( K_a ) of NH\u2084\u207a can be calculated using:<\/p>\n\n\n\n

    [
    K_a \\cdot K_b = K_w = 1.0 \\times 10^{-14}
    ]<\/p>\n\n\n\n

    Thus, the acid dissociation constant ( K_a ) of NH\u2084\u207a is:<\/p>\n\n\n\n

    [
    K_a = \\frac{1.0 \\times 10^{-14}}{1.8 \\times 10^{-5}} = 5.56 \\times 10^{-10}
    ]<\/p>\n\n\n\n

    Now, for the reaction:<\/p>\n\n\n\n

    [
    NH_4^+ + H_2O \\rightleftharpoons NH_3 + H_3O^+
    ]<\/p>\n\n\n\n

    We set up an ICE (Initial, Change, Equilibrium) table for NH\u2084\u207a dissociation:<\/p>\n\n\n\n

      \n
    • Initial concentration of NH\u2084\u207a = 0.03 M<\/li>\n\n\n\n
    • Change: Let the concentration of NH\u2084\u207a that dissociates be ( x ).<\/li>\n\n\n\n
    • At equilibrium, the concentrations will be: [NH\u2084\u207a] = 0.03 – x, [NH\u2083] = x, and [H\u2083O\u207a] = x.<\/li>\n<\/ul>\n\n\n\n

      The equilibrium expression for ( K_a ) is:<\/p>\n\n\n\n

      [
      K_a = \\frac{[NH_3][H_3O^+]}{[NH_4^+]} = \\frac{x \\cdot x}{0.03 – x}
      ]<\/p>\n\n\n\n

      Assuming ( x ) is small compared to 0.03, we can approximate the denominator as 0.03:<\/p>\n\n\n\n

      [
      K_a = \\frac{x^2}{0.03}
      ]<\/p>\n\n\n\n

      Now, solve for ( x ):<\/p>\n\n\n\n

      [
      x^2 = (5.56 \\times 10^{-10})(0.03)
      ]
      [
      x^2 = 1.668 \\times 10^{-11}
      ]
      [
      x = \\sqrt{1.668 \\times 10^{-11}} = 4.09 \\times 10^{-6} \\text{ M}
      ]<\/p>\n\n\n\n

      This value for ( x ) is the concentration of hydronium ions ([H_3O^+]). To find the pH:<\/p>\n\n\n\n

      [
      \\text{pH} = -\\log [H_3O^+] = -\\log (4.09 \\times 10^{-6})
      ]
      [
      \\text{pH} \\approx 5.39
      ]<\/p>\n\n\n\n

      Thus, the pH of a 0.03 M NH\u2084NO\u2083 solution is approximately 5.39<\/strong>, indicating that the solution is slightly acidic due to the presence of NH\u2084\u207a.<\/p>\n","protected":false},"excerpt":{"rendered":"

      Classify NH4NO3 as an acidic, basic, or neutral salt, and then answer the following question. What is the pH of 0.03 M NH,NO, (aq)? The Correct Answer and Explanation is : To classify NH\u2084NO\u2083 as an acidic, basic, or neutral salt, let’s break it down into its constituent ions: NH\u2084\u207a (ammonium ion) and NO\u2083\u207b (nitrate […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-189046","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189046","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=189046"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189046\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189046"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189046"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189046"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}