To solve this titration problem, we first need to understand the neutralization reaction between HBr (a strong acid) and LiOH (a strong base):HBr+LiOH\u2192LiBr+H2O\\text{HBr} + \\text{LiOH} \\rightarrow \\text{LiBr} + \\text{H}_2\\text{O}HBr+LiOH\u2192LiBr+H2\u200bO<\/p>\n\n\n\n
The stoichiometry of this reaction is 1:1, meaning one mole of HBr neutralizes one mole of LiOH.<\/p>\n\n\n\n
Step 1: Determine the moles of HBr used<\/h3>\n\n\n\n
We are told that 35.50 mL of 0.40 M HBr is required to reach the stoichiometric point. To find the moles of HBr, we use the formula:moles of HBr=Molarity\u00d7Volume (in L)\\text{moles of HBr} = \\text{Molarity} \\times \\text{Volume (in L)}moles of HBr=Molarity\u00d7Volume (in L)moles of HBr=0.40\u2009M\u00d735.50\u2009mL1000\u2009mL\/L\\text{moles of HBr} = 0.40 \\, \\text{M} \\times \\frac{35.50 \\, \\text{mL}}{1000 \\, \\text{mL\/L}}moles of HBr=0.40M\u00d71000mL\/L35.50mL\u200bmoles of HBr=0.40\u00d70.0355\u2009L=0.0142\u2009moles\\text{moles of HBr} = 0.40 \\times 0.0355 \\, \\text{L} = 0.0142 \\, \\text{moles}moles of HBr=0.40\u00d70.0355L=0.0142moles<\/p>\n\n\n\n
Step 2: Determine the moles of LiOH<\/h3>\n\n\n\n
At the stoichiometric point, the moles of HBr added are equal to the moles of LiOH in the sample. Therefore, moles of LiOH = moles of HBr = 0.0142 moles.<\/p>\n\n\n\n
Step 3: Calculate the concentration of LiOH<\/h3>\n\n\n\n
We are given that the volume of the LiOH sample is 25.00 mL, or 0.02500 L. To find the concentration of LiOH, we use the formula:Molarity of LiOH=moles of LiOHvolume of LiOH (in L)\\text{Molarity of LiOH} = \\frac{\\text{moles of LiOH}}{\\text{volume of LiOH (in L)}}Molarity of LiOH=volume of LiOH (in L)moles of LiOH\u200bMolarity of LiOH=0.0142\u2009moles0.02500\u2009L=0.568\u2009M\\text{Molarity of LiOH} = \\frac{0.0142 \\, \\text{moles}}{0.02500 \\, \\text{L}} = 0.568 \\, \\text{M}Molarity of LiOH=0.02500L0.0142moles\u200b=0.568M<\/p>\n\n\n\n
Step 4: Compare the calculated value to the given options<\/h3>\n\n\n\n
The calculated concentration of LiOH is 0.568 M, which is closest to 0.57 M<\/strong>.<\/p>\n\n\n\n Thus, the correct answer is 0.57 M<\/strong>.<\/p>\n\n\n\n This means the original concentration of LiOH in the sample is 0.57 M<\/strong>. The titration process helped us determine this by using the stoichiometric relationship between the acid and base in the neutralization reaction.<\/p>\n\n\n\n The stoichiometric point of a titration is reached when 35.50 mL 0.40 M HBr is added to a 25.00 mL sample of LiOH. The original [LiOH] is 0.57 M 0.014 M 0.28 M 0.024 M The Correct Answer and Explanation is: To solve this titration problem, we first need to understand the neutralization reaction between […]<\/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-248309","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248309","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=248309"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/248309\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=248309"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=248309"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=248309"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner8-200.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"