{"id":263348,"date":"2025-07-21T05:59:22","date_gmt":"2025-07-21T05:59:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=263348"},"modified":"2025-07-21T05:59:24","modified_gmt":"2025-07-21T05:59:24","slug":"given-the-balanced-equation-hbr-naoh-nabr-h2o-a-calculate-the-number-of-moles-of-hbr-in-100-0-ml-of-0-250-m-hbr","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/21\/given-the-balanced-equation-hbr-naoh-nabr-h2o-a-calculate-the-number-of-moles-of-hbr-in-100-0-ml-of-0-250-m-hbr\/","title":{"rendered":"Given the balanced equation: HBr + NaOH -> NaBr + H2O (a) Calculate the number of moles of HBr in 100.0 mL of 0.250 M HBr."},"content":{"rendered":"\n

Given the balanced equation: HBr + NaOH -> NaBr + H2O (a) Calculate the number of moles of HBr in 100.0 mL of 0.250 M HBr. Calculate the number of moles of NaOH in 100.0 mL of 0.250 M NaOH. When these two solutions are mixed, the acid and base should neutralize each other exactly. This means that all of the acid and base are completely used up; either one could be considered the “limiting reactant”. Starting with the moles of either the acid or base, calculate the number of moles of salt produced by the reaction. Using the moles of salt produced and the total volume in mL of the solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction:<\/p>\n\n\n\n

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

Let’s break this down step by step:<\/p>\n\n\n\n

Step 1: Calculate the moles of HBr in 100.0 mL of 0.250 M HBr.<\/h3>\n\n\n\n

The formula to calculate the number of moles is:moles=molarity\u00d7volume (in L)\\text{moles} = \\text{molarity} \\times \\text{volume (in L)}moles=molarity\u00d7volume (in L)<\/p>\n\n\n\n

For HBr:moles of HBr=0.250\u2009M\u00d70.1000\u2009L=0.0250\u2009mol\\text{moles of HBr} = 0.250 \\, \\text{M} \\times 0.1000 \\, \\text{L} = 0.0250 \\, \\text{mol}moles of HBr=0.250M\u00d70.1000L=0.0250mol<\/p>\n\n\n\n

Step 2: Calculate the moles of NaOH in 100.0 mL of 0.250 M NaOH.<\/h3>\n\n\n\n

Similarly, for NaOH:moles of NaOH=0.250\u2009M\u00d70.1000\u2009L=0.0250\u2009mol\\text{moles of NaOH} = 0.250 \\, \\text{M} \\times 0.1000 \\, \\text{L} = 0.0250 \\, \\text{mol}moles of NaOH=0.250M\u00d70.1000L=0.0250mol<\/p>\n\n\n\n

Step 3: Determine the moles of salt (NaBr) produced.<\/h3>\n\n\n\n

According to the balanced equation:HBr (aq)+NaOH (aq)\u2192NaBr (aq)+H2O (l)\\text{HBr (aq)} + \\text{NaOH (aq)} \\rightarrow \\text{NaBr (aq)} + \\text{H}_2\\text{O (l)}HBr (aq)+NaOH (aq)\u2192NaBr (aq)+H2\u200bO (l)<\/p>\n\n\n\n

The reaction occurs in a 1:1 molar ratio. Since you have equal moles of HBr and NaOH, both will neutralize each other completely, producing an equal number of moles of NaBr.<\/p>\n\n\n\n

Thus, the moles of NaBr produced will be:moles of NaBr=0.0250\u2009mol\\text{moles of NaBr} = 0.0250 \\, \\text{mol}moles of NaBr=0.0250mol<\/p>\n\n\n\n

Step 4: Calculate the molarity of the NaBr solution.<\/h3>\n\n\n\n

After mixing, the total volume of the solution will be the sum of the volumes of the HBr and NaOH solutions:total volume=100.0\u2009mL+100.0\u2009mL=200.0\u2009mL=0.2000\u2009L\\text{total volume} = 100.0 \\, \\text{mL} + 100.0 \\, \\text{mL} = 200.0 \\, \\text{mL} = 0.2000 \\, \\text{L}total volume=100.0mL+100.0mL=200.0mL=0.2000L<\/p>\n\n\n\n

Now, to calculate the molarity of NaBr, use the formula:molarity of NaBr=moles of NaBrtotal volume (in L)\\text{molarity of NaBr} = \\frac{\\text{moles of NaBr}}{\\text{total volume (in L)}}molarity of NaBr=total volume (in L)moles of NaBr\u200bmolarity of NaBr=0.0250\u2009mol0.2000\u2009L=0.125\u2009M\\text{molarity of NaBr} = \\frac{0.0250 \\, \\text{mol}}{0.2000 \\, \\text{L}} = 0.125 \\, \\text{M}molarity of NaBr=0.2000L0.0250mol\u200b=0.125M<\/p>\n\n\n\n

Final Answer:<\/h3>\n\n\n\n
    \n
  • Moles of HBr: 0.0250 mol<\/strong><\/li>\n\n\n\n
  • Moles of NaOH: 0.0250 mol<\/strong><\/li>\n\n\n\n
  • Moles of NaBr produced: 0.0250 mol<\/strong><\/li>\n\n\n\n
  • Molarity of NaBr solution: 0.125 M<\/strong><\/li>\n<\/ul>\n\n\n\n

    This shows that when equal volumes of equal molar concentrations of HBr and NaOH are mixed, they completely neutralize each other to form NaBr, and the molarity of the resulting salt solution is half of the original molarity of the acid or base.<\/p>\n\n\n\n

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

    Given the balanced equation: HBr + NaOH -> NaBr + H2O (a) Calculate the number of moles of HBr in 100.0 mL of 0.250 M HBr. Calculate the number of moles of NaOH in 100.0 mL of 0.250 M NaOH. When these two solutions are mixed, the acid and base should neutralize each other exactly. […]<\/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-263348","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/263348","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=263348"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/263348\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=263348"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=263348"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=263348"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}