{"id":221018,"date":"2025-05-29T05:57:48","date_gmt":"2025-05-29T05:57:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221018"},"modified":"2025-05-29T05:57:50","modified_gmt":"2025-05-29T05:57:50","slug":"what-is-the-molar-concentration-of-nh3-in-this-solution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/what-is-the-molar-concentration-of-nh3-in-this-solution\/","title":{"rendered":"What is the molar concentration of NH3\u00a0in this solution"},"content":{"rendered":"\n

A concentrated solution of aqueous ammonia is 28.0% w\/w NH3<\/sub> and has a density of 0.899 g\/mL. What is the molar concentration of NH3<\/sub> in this solution?<\/p>\n\n\n\n

What is pNa for a solution of 1.76 \u00d7 10\u20133<\/sup> M Na3<\/sub>PO4<\/sub>?<\/p>\n\n\n\n

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

1. Molar concentration of NH\u2083 in 28.0% w\/w aqueous ammonia<\/strong><\/h3>\n\n\n\n

Given:<\/strong><\/p>\n\n\n\n

    \n
  • 28.0% w\/w NH\u2083 \u2192 28.0 g NH\u2083 in 100 g solution<\/li>\n\n\n\n
  • Density of solution = 0.899 g\/mL<\/li>\n\n\n\n
  • Molar mass of NH\u2083 = 17.03 g\/mol<\/li>\n<\/ul>\n\n\n\n

    Step 1: Convert total mass of solution to volume<\/strong>
    Mass of solution = 100 g
    Volume = mass \/ density = 100 g \/ 0.899 g\/mL \u2248 111.23 mL = 0.11123 L<\/p>\n\n\n\n

    Step 2: Calculate moles of NH\u2083<\/strong>
    Moles NH\u2083 = 28.0 g \/ 17.03 g\/mol \u2248 1.644 mol<\/p>\n\n\n\n

    Step 3: Calculate molar concentration (mol\/L)<\/strong>
    Molarity = moles \/ volume (L) = 1.644 mol \/ 0.11123 L \u2248 14.78 M<\/strong><\/p>\n\n\n\n

    \u2705 Answer: 14.78 M NH\u2083<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    2. pNa of 1.76 \u00d7 10\u207b\u00b3 M Na\u2083PO\u2084 solution<\/strong><\/h3>\n\n\n\n

    Step 1: Understand dissociation<\/strong>
    Na\u2083PO\u2084 dissociates completely in water:
    Na\u2083PO\u2084 \u2192 3Na\u207a + PO\u2084\u00b3\u207b<\/p>\n\n\n\n

    So, for every 1 mole of Na\u2083PO\u2084, we get 3 moles of Na\u207a.<\/p>\n\n\n\n

    Step 2: Calculate [Na\u207a]<\/strong>
    [Na\u2083PO\u2084] = 1.76 \u00d7 10\u207b\u00b3 M
    \u21d2 [Na\u207a] = 3 \u00d7 (1.76 \u00d7 10\u207b\u00b3) = 5.28 \u00d7 10\u207b\u00b3 M<\/p>\n\n\n\n

    Step 3: Calculate pNa<\/strong>
    pNa = \u2013log\u2081\u2080[Na\u207a] = \u2013log\u2081\u2080(5.28 \u00d7 10\u207b\u00b3) \u2248 2.28<\/p>\n\n\n\n

    \u2705 Answer: pNa = 2.28<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Explanation <\/strong><\/h3>\n\n\n\n

    The molar concentration<\/strong> (or molarity) of a solute in a solution refers to the number of moles of solute per liter of solution. For aqueous ammonia, a 28.0% w\/w solution indicates 28.0 g of NH\u2083 is present in every 100 g of the total solution. Given the density of the solution is 0.899 g\/mL, we use this to convert the mass of the solution to volume, which allows us to express molarity in mol\/L. By dividing the mass of NH\u2083 by its molar mass (17.03 g\/mol), we obtain the number of moles of ammonia. Dividing this by the calculated volume in liters yields a concentration of 14.78 M<\/strong>, which is highly concentrated, as expected for industrial-grade aqueous ammonia.<\/p>\n\n\n\n

    On the other hand, pNa<\/strong> is a logarithmic scale similar to pH or pOH, but it refers to the sodium ion concentration. In a solution of sodium phosphate (Na\u2083PO\u2084), the salt dissociates completely into 3 sodium ions and one phosphate ion. This stoichiometry is crucial because it triples the effective sodium ion concentration compared to the initial Na\u2083PO\u2084 concentration. Multiplying 1.76 \u00d7 10\u207b\u00b3 M by 3 gives the sodium ion concentration of 5.28 \u00d7 10\u207b\u00b3 M. Applying the negative logarithm base 10, the pNa is calculated to be approximately 2.28<\/strong>. Low pNa values indicate high concentrations of Na\u207a, which can be important in contexts like electrochemistry, biochemistry, or ion-selective electrode measurements.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    A concentrated solution of aqueous ammonia is 28.0% w\/w NH3 and has a density of 0.899 g\/mL. What is the molar concentration of NH3 in this solution? What is pNa for a solution of 1.76 \u00d7 10\u20133 M Na3PO4? The Correct Answer and Explanation is: 1. Molar concentration of NH\u2083 in 28.0% w\/w aqueous ammonia Given: Step 1: […]<\/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-221018","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221018","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=221018"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221018\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221018"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221018"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221018"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}