{"id":47479,"date":"2025-07-02T12:26:08","date_gmt":"2025-07-02T12:26:08","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=47479"},"modified":"2025-07-02T12:26:10","modified_gmt":"2025-07-02T12:26:10","slug":"ammonia-nh-is-an-example-of-a-bronsted-lowry-base","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/ammonia-nh-is-an-example-of-a-bronsted-lowry-base\/","title":{"rendered":"Ammonia (NH) is an example of a Br\u00f8nsted-Lowry Base."},"content":{"rendered":"\n<p>. Ammonia (NH<br>) is an example of a Br\u00f8nsted-Lowry Base. i. Define the Br\u00f8nsted-Lowry acid-base theory. ii. What&#8217;s the pH of a solution of ammonia that has a concentration of 0.335 M? The K<br>of ammonia is<br>. First complete the ICE chart. HA H<br>O<br>A<br>Initial Change Equilibrium iii. Calculate the pH. Please show all work.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Here is the solution to the chemistry problem.<\/p>\n\n\n\n<p><strong>11. Ammonia (NH\u2083) is an example of a Br\u00f8nsted-Lowry Base.<\/strong><\/p>\n\n\n\n<p><strong>i. Define the Br\u00f8nsted-Lowry acid-base theory.<\/strong><br>The Br\u00f8nsted-Lowry theory defines acids and bases based on the transfer of protons (H\u207a ions).<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A\u00a0<strong>Br\u00f8nsted-Lowry acid<\/strong>\u00a0is a substance that donates a proton (H\u207a).<\/li>\n\n\n\n<li>A\u00a0<strong>Br\u00f8nsted-Lowry base<\/strong>\u00a0is a substance that accepts a proton (H\u207a).<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>ii. What&#8217;s the pH of a solution of ammonia that has a concentration of 0.335 M? The K\u208b of ammonia is 1.8 \u00d7 10\u207b\u2075. First complete the ICE chart.<\/strong><\/p>\n\n\n\n<p>The problem involves ammonia (NH\u2083), a weak base, reacting with water. The chemical equation is:<br>NH\u2083(aq) + H\u2082O(l) \u21cc NH\u2084\u207a(aq) + OH\u207b(aq)<\/p>\n\n\n\n<p><em>Note: The ICE chart provided in the question has column headers (HA, H\u2083O\u207a, A\u207b) that are appropriate for a generic acid dissociation. For this problem about the base ammonia, the correct species are the reactant (NH\u2083), the conjugate acid (NH\u2084\u207a), and the hydroxide ion (OH\u207b). The chart below is completed with the correct values for the ammonia reaction, corresponding to the species NH\u2083, NH\u2084\u207a, and OH\u207b.<\/em><\/p>\n\n\n\n<p>Here is the completed ICE chart for the reaction of ammonia in water:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><tbody><tr><td><\/td><td><strong>NH\u2083<\/strong><\/td><td><strong>NH\u2084\u207a<\/strong><\/td><td><strong>OH\u207b<\/strong><\/td><\/tr><tr><td><strong>Initial<\/strong><\/td><td>0.335<\/td><td>0<\/td><td>0<\/td><\/tr><tr><td><strong>Change<\/strong><\/td><td>-x<\/td><td>+x<\/td><td>+x<\/td><\/tr><tr><td><strong>Equilibrium<\/strong><\/td><td>0.335 &#8211; x<\/td><td>x<\/td><td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>iii. Calculate the pH. Please show all work.<\/strong><\/p>\n\n\n\n<p><strong>Step 1: Write the base dissociation constant (K\u208b) expression.<\/strong><br>The expression for K\u208b is based on the equilibrium concentrations from the ICE chart.<br>K\u208b = [NH\u2084\u207a][OH\u207b] \/ [NH\u2083]<\/p>\n\n\n\n<p><strong>Step 2: Substitute the equilibrium values into the K\u208b expression.<\/strong><br>1.8 \u00d7 10\u207b\u2075 = (x)(x) \/ (0.335 &#8211; x)<\/p>\n\n\n\n<p><strong>Step 3: Simplify and solve for x.<\/strong><br>Since K\u208b is very small, we can assume that x will be much smaller than the initial concentration of 0.335 M. Therefore, we can approximate (0.335 &#8211; x) \u2248 0.335.<br>(To check this assumption: [NH\u2083]initial \/ K\u208b = 0.335 \/ (1.8 \u00d7 10\u207b\u2075) \u2248 18611, which is much greater than 100, so the approximation is valid).<\/p>\n\n\n\n<p>1.8 \u00d7 10\u207b\u2075 = x\u00b2 \/ 0.335<br>x\u00b2 = (1.8 \u00d7 10\u207b\u2075) \u00d7 0.335<br>x\u00b2 = 6.03 \u00d7 10\u207b\u2076<br>x = \u221a(6.03 \u00d7 10\u207b\u2076)<br>x = 2.456 \u00d7 10\u207b\u00b3<\/p>\n\n\n\n<p><strong>Step 4: Calculate the pOH.<\/strong><br>From the ICE chart, x represents the equilibrium concentration of hydroxide ions, [OH\u207b].<br>[OH\u207b] = 2.456 \u00d7 10\u207b\u00b3 M<\/p>\n\n\n\n<p>pOH = -log[OH\u207b]<br>pOH = -log(2.456 \u00d7 10\u207b\u00b3)<br>pOH \u2248 2.61<\/p>\n\n\n\n<p><strong>Step 5: Calculate the pH.<\/strong><br>The relationship between pH and pOH at 25\u00b0C is pH + pOH = 14.<br>pH = 14 &#8211; pOH<br>pH = 14 &#8211; 2.61<br><strong>pH = 11.39<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-245.jpeg\" alt=\"\" class=\"wp-image-47483\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-245.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-245-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-245-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-245-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>. Ammonia (NH) is an example of a Br\u00f8nsted-Lowry Base. i. Define the Br\u00f8nsted-Lowry acid-base theory. ii. What&#8217;s the pH of a solution of ammonia that has a concentration of 0.335 M? The Kof ammonia is. First complete the ICE chart. HA HOAInitial Change Equilibrium iii. Calculate the pH. Please show all work. The Correct [&hellip;]<\/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-47479","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47479","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=47479"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47479\/revisions"}],"predecessor-version":[{"id":47484,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/47479\/revisions\/47484"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=47479"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=47479"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=47479"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}