{"id":221022,"date":"2025-05-29T06:02:58","date_gmt":"2025-05-29T06:02:58","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221022"},"modified":"2025-05-29T06:03:01","modified_gmt":"2025-05-29T06:03:01","slug":"what-is-the-molar-concentration-of-nh3-in-this-solution-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/what-is-the-molar-concentration-of-nh3-in-this-solution-2\/","title":{"rendered":"What is the molar concentration of NH3 in this solution"},"content":{"rendered":"\n

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?<\/p>\n\n\n\n

What is pNa for a solution of 1.76 \u00d7 10\u20133 M Na3PO4?<\/p>\n\n\n\n

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

Part 1: Molar concentration of NH\u2083<\/strong><\/h3>\n\n\n\n

We are given:<\/p>\n\n\n\n

    \n
  • 28.0% w\/w NH\u2083 \u2192 28.0 g NH\u2083 per 100 g of solution<\/li>\n\n\n\n
  • Density = 0.899 g\/mL<\/li>\n<\/ul>\n\n\n\n

    Step 1: Find the mass of the solution per liter<\/strong><\/p>\n\n\n\n

    Since density = mass\/volume: Mass of 1 L solution=0.899\u2009g\/mL\u00d71000\u2009mL=899\u2009g\\text{Mass of 1 L solution} = 0.899 \\, \\text{g\/mL} \\times 1000 \\, \\text{mL} = 899 \\, \\text{g}<\/p>\n\n\n\n

    Step 2: Find mass of NH\u2083 in 1 L solution<\/strong><\/p>\n\n\n\n

    28.0% of 899 g is NH\u2083: Mass of NH\u2083=0.28\u00d7899=251.72\u2009g\\text{Mass of NH\u2083} = 0.28 \\times 899 = 251.72 \\, \\text{g}<\/p>\n\n\n\n

    Step 3: Convert NH\u2083 mass to moles<\/strong><\/p>\n\n\n\n

    Molar mass of NH\u2083 = 17.03 g\/mol Moles of NH\u2083=251.7217.03=14.78\u2009mol\\text{Moles of NH\u2083} = \\frac{251.72}{17.03} = 14.78 \\, \\text{mol}<\/p>\n\n\n\n

    Answer:<\/strong> Molarity=14.78\u2009mol1.00\u2009L=14.78\u2009M\\text{Molarity} = \\frac{14.78 \\, \\text{mol}}{1.00 \\, \\text{L}} = \\boxed{14.78 \\, \\text{M}}<\/p>\n\n\n\n

    \n\n\n\n

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

    Step 1: Dissociation of Na\u2083PO\u2084<\/strong> Na3PO4\u21923Na++PO43\u2212\\text{Na}_3\\text{PO}_4 \\rightarrow 3\\text{Na}^+ + \\text{PO}_4^{3-}<\/p>\n\n\n\n

    For every 1 mol of Na\u2083PO\u2084, 3 mol of Na\u207a are produced.<\/p>\n\n\n\n

    Step 2: Calculate [Na\u207a]<\/strong> [Na+]=3\u00d7(1.76\u00d710\u22123)=5.28\u00d710\u22123\u2009M[\\text{Na}^+] = 3 \\times (1.76 \\times 10^{-3}) = 5.28 \\times 10^{-3} \\, \\text{M}<\/p>\n\n\n\n

    Step 3: Calculate pNa<\/strong><\/p>\n\n\n\n

    pNa is defined as: pNa=\u2212log\u206110[Na+]\\text{pNa} = -\\log_{10}[\\text{Na}^+] pNa=\u2212log\u206110(5.28\u00d710\u22123)\u22482.28\\text{pNa} = -\\log_{10}(5.28 \\times 10^{-3}) \\approx \\boxed{2.28}<\/p>\n\n\n\n

    \n\n\n\n

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

    In chemistry, concentration calculations and logarithmic expressions are crucial for understanding solution properties. For the ammonia solution, “28.0% w\/w” means that 28 grams of ammonia are present per 100 grams of solution. To determine the molarity (mol\/L), we must first find the total mass of one liter of solution using its density. With a density of 0.899 g\/mL, one liter (1000 mL) weighs 899 grams. Using the 28.0% ratio, the actual mass of ammonia in that volume is 251.72 grams. Dividing this mass by the molar mass of ammonia (17.03 g\/mol) yields the number of moles: 14.78 mol. Since this is in one liter, the molarity is 14.78 M. This value indicates a highly concentrated basic solution, typical for industrial-strength ammonia.<\/p>\n\n\n\n

    For pNa, we use the concept similar to pH. Sodium phosphate (Na\u2083PO\u2084) dissociates in water to give 3 Na\u207a ions for every formula unit. Therefore, a 1.76 \u00d7 10\u207b\u00b3 M solution of Na\u2083PO\u2084 releases three times as many Na\u207a ions: 5.28 \u00d7 10\u207b\u00b3 M. The pNa is calculated as the negative logarithm of the sodium ion concentration: \u2212log\u2081\u2080[Na\u207a]. This gives a pNa of about 2.28, meaning the concentration of sodium ions is relatively high. This calculation is important in fields like electrochemistry or biochemistry, where specific ion concentrations affect cell behavior or reaction mechanisms. Understanding how to go from percentage concentrations and stoichiometric relationships to molarity and p-functions (like pH, pKa, or pNa) is fundamental in both lab and theoretical chemistry.<\/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: Part 1: Molar concentration of NH\u2083 We are given: Step […]<\/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-221022","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221022","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=221022"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221022\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221022"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221022"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221022"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}