Answer Summary<\/strong><\/h3>\n\n\n\n\n- After adding 10.0 mL of NaOH to 20.0 mL of 0.200 M HBrO<\/strong>,
pH \u2248 8.66<\/strong><\/li>\n\n\n\n- At the equivalence point of titrating 25.0 mL of 0.600 M HBrO with 0.400 M NaOH<\/strong>,
pH \u2248 10.34<\/strong><\/li>\n<\/ol>\n\n\n\n
\n\n\n\nDetailed Explanation<\/strong><\/h3>\n\n\n\nHypobromous acid (HBrO) is a weak acid<\/strong>, and NaOH is a strong base<\/strong>. In acid-base titrations involving a weak acid and a strong base, the pH depends on the stage of the titration.<\/p>\n\n\n\n
\n\n\n\n1. After Adding 10.0 mL of NaOH to 20.0 mL of 0.200 M HBrO<\/strong><\/h4>\n\n\n\nStep 1: Moles of Reactants<\/strong><\/p>\n\n\n\n\n- Moles of HBrO = 0.0200 L \u00d7 0.200 mol\/L = 0.00400 mol<\/li>\n\n\n\n
- Moles of NaOH = 0.0100 L \u00d7 0.250 mol\/L = 0.00250 mol<\/li>\n<\/ul>\n\n\n\n
Step 2: Reaction<\/strong>
HBrO+OH\u2212\u2192BrO\u2212+H2O\\text{HBrO} + \\text{OH}^- \\rightarrow \\text{BrO}^- + H_2O<\/p>\n\n\n\n\n- NaOH reacts with HBrO to form BrO\u207b.<\/li>\n\n\n\n
- Moles of HBrO remaining = 0.00400 \u2013 0.00250 = 0.00150 mol<\/li>\n\n\n\n
- Moles of BrO\u207b formed = 0.00250 mol<\/li>\n\n\n\n
- Total volume = 20.0 + 10.0 = 30.0 mL = 0.0300 L<\/li>\n<\/ul>\n\n\n\n
Step 3: Use Henderson-Hasselbalch Equation<\/strong> pKa=\u2212log\u2061(2.3\u00d710\u22129)\u22488.64\\text{pKa} = -\\log(2.3 \\times 10^{-9}) \u2248 8.64 pH=pKa+log\u2061([BrO\u2212][HBrO])=8.64+log\u2061(0.002500.00150)=8.64+log\u2061(1.6667)\u22488.64+0.22=8.66\\text{pH} = pKa + \\log\\left( \\frac{[\\text{BrO}^-]}{[\\text{HBrO}]} \\right) = 8.64 + \\log\\left( \\frac{0.00250}{0.00150} \\right) = 8.64 + \\log(1.6667) \u2248 8.64 + 0.22 = \\boxed{8.66}<\/p>\n\n\n\n
\n\n\n\n2. At the Equivalence Point of Titrating 25.0 mL of 0.600 M HBrO with 0.400 M NaOH<\/strong><\/h4>\n\n\n\nStep 1: Moles of HBrO =<\/strong> 0.025 L \u00d7 0.600 mol\/L = 0.0150 mol
Since NaOH is in stoichiometric proportion, 0.0150 mol is required.<\/p>\n\n\n\nVolume of NaOH needed:<\/strong> V=0.0150 mol0.400 mol\/L=0.0375 L=37.5 mLV = \\frac{0.0150\\ \\text{mol}}{0.400\\ \\text{mol\/L}} = 0.0375\\ \\text{L} = 37.5\\ \\text{mL}<\/p>\n\n\n\nTotal volume = 25.0 + 37.5 = 62.5 mL = 0.0625 L<\/p>\n\n\n\n
At equivalence, all HBrO is converted to BrO\u207b (a weak base).<\/p>\n\n\n\n
[BrO\u207b] =<\/strong> 0.0150 mol \/ 0.0625 L = 0.240 M<\/p>\n\n\n\nNow calculate pH using base hydrolysis: Kb=KwKa=1.0\u00d710\u2212142.3\u00d710\u22129=4.35\u00d710\u22126\\text{Kb} = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{2.3 \\times 10^{-9}} = 4.35 \\times 10^{-6} [OH\u2212]=Kb\u00d7[BrO\u2212]=4.35\u00d710\u22126\u00d70.240\u22481.044\u00d710\u22126\u22481.02\u00d710\u22123[OH^-] = \\sqrt{Kb \\times [BrO^-]} = \\sqrt{4.35 \\times 10^{-6} \\times 0.240} \u2248 \\sqrt{1.044 \\times 10^{-6}} \u2248 1.02 \\times 10^{-3} pOH=\u2212log\u2061(1.02\u00d710\u22123)\u22482.99\u21d2pH=14\u22122.99=10.34\\text{pOH} = -\\log(1.02 \\times 10^{-3}) \u2248 2.99 \\quad \\Rightarrow \\quad \\text{pH} = 14 – 2.99 = \\boxed{10.34}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/h3>\n\n\n\n\n- pH after 10.0 mL NaOH added<\/strong>: 8.66<\/strong><\/li>\n\n\n\n
- pH at equivalence point<\/strong>: 10.34<\/strong><\/li>\n<\/ul>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner8-16.jpeg\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"20.0 ml of 0.200M hypobromous acid, HBrO, is titrated with 0.250M sodium hydroxide, NaOH. Calculate the pH of the solution after the addition of 10.0 ml of NaOH solution. Then Determine 25.0 ml of 0.600M hypobromous acid, HBrO, is titrated with 0.400M sodium hydroxide, NaOH. Calculate the pH of the solution at the equivalence point […]<\/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-219472","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219472","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=219472"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219472\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
pH \u2248 8.66<\/strong><\/li>\n\n\n\n
pH \u2248 10.34<\/strong><\/li>\n<\/ol>\n\n\n\n
\n\n\n\n
Detailed Explanation<\/strong><\/h3>\n\n\n\nHypobromous acid (HBrO) is a weak acid<\/strong>, and NaOH is a strong base<\/strong>. In acid-base titrations involving a weak acid and a strong base, the pH depends on the stage of the titration.<\/p>\n\n\n\n
\n\n\n\n1. After Adding 10.0 mL of NaOH to 20.0 mL of 0.200 M HBrO<\/strong><\/h4>\n\n\n\nStep 1: Moles of Reactants<\/strong><\/p>\n\n\n\n\n- Moles of HBrO = 0.0200 L \u00d7 0.200 mol\/L = 0.00400 mol<\/li>\n\n\n\n
- Moles of NaOH = 0.0100 L \u00d7 0.250 mol\/L = 0.00250 mol<\/li>\n<\/ul>\n\n\n\n
Step 2: Reaction<\/strong>
HBrO+OH\u2212\u2192BrO\u2212+H2O\\text{HBrO} + \\text{OH}^- \\rightarrow \\text{BrO}^- + H_2O<\/p>\n\n\n\n\n- NaOH reacts with HBrO to form BrO\u207b.<\/li>\n\n\n\n
- Moles of HBrO remaining = 0.00400 \u2013 0.00250 = 0.00150 mol<\/li>\n\n\n\n
- Moles of BrO\u207b formed = 0.00250 mol<\/li>\n\n\n\n
- Total volume = 20.0 + 10.0 = 30.0 mL = 0.0300 L<\/li>\n<\/ul>\n\n\n\n
Step 3: Use Henderson-Hasselbalch Equation<\/strong> pKa=\u2212log\u2061(2.3\u00d710\u22129)\u22488.64\\text{pKa} = -\\log(2.3 \\times 10^{-9}) \u2248 8.64 pH=pKa+log\u2061([BrO\u2212][HBrO])=8.64+log\u2061(0.002500.00150)=8.64+log\u2061(1.6667)\u22488.64+0.22=8.66\\text{pH} = pKa + \\log\\left( \\frac{[\\text{BrO}^-]}{[\\text{HBrO}]} \\right) = 8.64 + \\log\\left( \\frac{0.00250}{0.00150} \\right) = 8.64 + \\log(1.6667) \u2248 8.64 + 0.22 = \\boxed{8.66}<\/p>\n\n\n\n
\n\n\n\n2. At the Equivalence Point of Titrating 25.0 mL of 0.600 M HBrO with 0.400 M NaOH<\/strong><\/h4>\n\n\n\nStep 1: Moles of HBrO =<\/strong> 0.025 L \u00d7 0.600 mol\/L = 0.0150 mol
Since NaOH is in stoichiometric proportion, 0.0150 mol is required.<\/p>\n\n\n\nVolume of NaOH needed:<\/strong> V=0.0150 mol0.400 mol\/L=0.0375 L=37.5 mLV = \\frac{0.0150\\ \\text{mol}}{0.400\\ \\text{mol\/L}} = 0.0375\\ \\text{L} = 37.5\\ \\text{mL}<\/p>\n\n\n\nTotal volume = 25.0 + 37.5 = 62.5 mL = 0.0625 L<\/p>\n\n\n\n
At equivalence, all HBrO is converted to BrO\u207b (a weak base).<\/p>\n\n\n\n
[BrO\u207b] =<\/strong> 0.0150 mol \/ 0.0625 L = 0.240 M<\/p>\n\n\n\nNow calculate pH using base hydrolysis: Kb=KwKa=1.0\u00d710\u2212142.3\u00d710\u22129=4.35\u00d710\u22126\\text{Kb} = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{2.3 \\times 10^{-9}} = 4.35 \\times 10^{-6} [OH\u2212]=Kb\u00d7[BrO\u2212]=4.35\u00d710\u22126\u00d70.240\u22481.044\u00d710\u22126\u22481.02\u00d710\u22123[OH^-] = \\sqrt{Kb \\times [BrO^-]} = \\sqrt{4.35 \\times 10^{-6} \\times 0.240} \u2248 \\sqrt{1.044 \\times 10^{-6}} \u2248 1.02 \\times 10^{-3} pOH=\u2212log\u2061(1.02\u00d710\u22123)\u22482.99\u21d2pH=14\u22122.99=10.34\\text{pOH} = -\\log(1.02 \\times 10^{-3}) \u2248 2.99 \\quad \\Rightarrow \\quad \\text{pH} = 14 – 2.99 = \\boxed{10.34}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/h3>\n\n\n\n\n- pH after 10.0 mL NaOH added<\/strong>: 8.66<\/strong><\/li>\n\n\n\n
- pH at equivalence point<\/strong>: 10.34<\/strong><\/li>\n<\/ul>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"20.0 ml of 0.200M hypobromous acid, HBrO, is titrated with 0.250M sodium hydroxide, NaOH. Calculate the pH of the solution after the addition of 10.0 ml of NaOH solution. Then Determine 25.0 ml of 0.600M hypobromous acid, HBrO, is titrated with 0.400M sodium hydroxide, NaOH. Calculate the pH of the solution at the equivalence point […]<\/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-219472","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219472","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=219472"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219472\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n
1. After Adding 10.0 mL of NaOH to 20.0 mL of 0.200 M HBrO<\/strong><\/h4>\n\n\n\nStep 1: Moles of Reactants<\/strong><\/p>\n\n\n\n\n- Moles of HBrO = 0.0200 L \u00d7 0.200 mol\/L = 0.00400 mol<\/li>\n\n\n\n
- Moles of NaOH = 0.0100 L \u00d7 0.250 mol\/L = 0.00250 mol<\/li>\n<\/ul>\n\n\n\n
Step 2: Reaction<\/strong>
HBrO+OH\u2212\u2192BrO\u2212+H2O\\text{HBrO} + \\text{OH}^- \\rightarrow \\text{BrO}^- + H_2O<\/p>\n\n\n\n\n- NaOH reacts with HBrO to form BrO\u207b.<\/li>\n\n\n\n
- Moles of HBrO remaining = 0.00400 \u2013 0.00250 = 0.00150 mol<\/li>\n\n\n\n
- Moles of BrO\u207b formed = 0.00250 mol<\/li>\n\n\n\n
- Total volume = 20.0 + 10.0 = 30.0 mL = 0.0300 L<\/li>\n<\/ul>\n\n\n\n
Step 3: Use Henderson-Hasselbalch Equation<\/strong> pKa=\u2212log\u2061(2.3\u00d710\u22129)\u22488.64\\text{pKa} = -\\log(2.3 \\times 10^{-9}) \u2248 8.64 pH=pKa+log\u2061([BrO\u2212][HBrO])=8.64+log\u2061(0.002500.00150)=8.64+log\u2061(1.6667)\u22488.64+0.22=8.66\\text{pH} = pKa + \\log\\left( \\frac{[\\text{BrO}^-]}{[\\text{HBrO}]} \\right) = 8.64 + \\log\\left( \\frac{0.00250}{0.00150} \\right) = 8.64 + \\log(1.6667) \u2248 8.64 + 0.22 = \\boxed{8.66}<\/p>\n\n\n\n
\n\n\n\n2. At the Equivalence Point of Titrating 25.0 mL of 0.600 M HBrO with 0.400 M NaOH<\/strong><\/h4>\n\n\n\nStep 1: Moles of HBrO =<\/strong> 0.025 L \u00d7 0.600 mol\/L = 0.0150 mol
Since NaOH is in stoichiometric proportion, 0.0150 mol is required.<\/p>\n\n\n\nVolume of NaOH needed:<\/strong> V=0.0150 mol0.400 mol\/L=0.0375 L=37.5 mLV = \\frac{0.0150\\ \\text{mol}}{0.400\\ \\text{mol\/L}} = 0.0375\\ \\text{L} = 37.5\\ \\text{mL}<\/p>\n\n\n\nTotal volume = 25.0 + 37.5 = 62.5 mL = 0.0625 L<\/p>\n\n\n\n
At equivalence, all HBrO is converted to BrO\u207b (a weak base).<\/p>\n\n\n\n
[BrO\u207b] =<\/strong> 0.0150 mol \/ 0.0625 L = 0.240 M<\/p>\n\n\n\nNow calculate pH using base hydrolysis: Kb=KwKa=1.0\u00d710\u2212142.3\u00d710\u22129=4.35\u00d710\u22126\\text{Kb} = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{2.3 \\times 10^{-9}} = 4.35 \\times 10^{-6} [OH\u2212]=Kb\u00d7[BrO\u2212]=4.35\u00d710\u22126\u00d70.240\u22481.044\u00d710\u22126\u22481.02\u00d710\u22123[OH^-] = \\sqrt{Kb \\times [BrO^-]} = \\sqrt{4.35 \\times 10^{-6} \\times 0.240} \u2248 \\sqrt{1.044 \\times 10^{-6}} \u2248 1.02 \\times 10^{-3} pOH=\u2212log\u2061(1.02\u00d710\u22123)\u22482.99\u21d2pH=14\u22122.99=10.34\\text{pOH} = -\\log(1.02 \\times 10^{-3}) \u2248 2.99 \\quad \\Rightarrow \\quad \\text{pH} = 14 – 2.99 = \\boxed{10.34}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/h3>\n\n\n\n\n- pH after 10.0 mL NaOH added<\/strong>: 8.66<\/strong><\/li>\n\n\n\n
- pH at equivalence point<\/strong>: 10.34<\/strong><\/li>\n<\/ul>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"20.0 ml of 0.200M hypobromous acid, HBrO, is titrated with 0.250M sodium hydroxide, NaOH. Calculate the pH of the solution after the addition of 10.0 ml of NaOH solution. Then Determine 25.0 ml of 0.600M hypobromous acid, HBrO, is titrated with 0.400M sodium hydroxide, NaOH. Calculate the pH of the solution at the equivalence point […]<\/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-219472","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219472","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=219472"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219472\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- \n
- Moles of HBrO = 0.0200 L \u00d7 0.200 mol\/L = 0.00400 mol<\/li>\n\n\n\n
- Moles of NaOH = 0.0100 L \u00d7 0.250 mol\/L = 0.00250 mol<\/li>\n<\/ul>\n\n\n\n
Step 2: Reaction<\/strong>
HBrO+OH\u2212\u2192BrO\u2212+H2O\\text{HBrO} + \\text{OH}^- \\rightarrow \\text{BrO}^- + H_2O<\/p>\n\n\n\n- \n
- NaOH reacts with HBrO to form BrO\u207b.<\/li>\n\n\n\n
- Moles of HBrO remaining = 0.00400 \u2013 0.00250 = 0.00150 mol<\/li>\n\n\n\n
- Moles of BrO\u207b formed = 0.00250 mol<\/li>\n\n\n\n
- Total volume = 20.0 + 10.0 = 30.0 mL = 0.0300 L<\/li>\n<\/ul>\n\n\n\n
Step 3: Use Henderson-Hasselbalch Equation<\/strong> pKa=\u2212log\u2061(2.3\u00d710\u22129)\u22488.64\\text{pKa} = -\\log(2.3 \\times 10^{-9}) \u2248 8.64 pH=pKa+log\u2061([BrO\u2212][HBrO])=8.64+log\u2061(0.002500.00150)=8.64+log\u2061(1.6667)\u22488.64+0.22=8.66\\text{pH} = pKa + \\log\\left( \\frac{[\\text{BrO}^-]}{[\\text{HBrO}]} \\right) = 8.64 + \\log\\left( \\frac{0.00250}{0.00150} \\right) = 8.64 + \\log(1.6667) \u2248 8.64 + 0.22 = \\boxed{8.66}<\/p>\n\n\n\n
\n\n\n\n2. At the Equivalence Point of Titrating 25.0 mL of 0.600 M HBrO with 0.400 M NaOH<\/strong><\/h4>\n\n\n\n
Step 1: Moles of HBrO =<\/strong> 0.025 L \u00d7 0.600 mol\/L = 0.0150 mol
Since NaOH is in stoichiometric proportion, 0.0150 mol is required.<\/p>\n\n\n\nVolume of NaOH needed:<\/strong> V=0.0150 mol0.400 mol\/L=0.0375 L=37.5 mLV = \\frac{0.0150\\ \\text{mol}}{0.400\\ \\text{mol\/L}} = 0.0375\\ \\text{L} = 37.5\\ \\text{mL}<\/p>\n\n\n\n
Total volume = 25.0 + 37.5 = 62.5 mL = 0.0625 L<\/p>\n\n\n\n
At equivalence, all HBrO is converted to BrO\u207b (a weak base).<\/p>\n\n\n\n
[BrO\u207b] =<\/strong> 0.0150 mol \/ 0.0625 L = 0.240 M<\/p>\n\n\n\n
Now calculate pH using base hydrolysis: Kb=KwKa=1.0\u00d710\u2212142.3\u00d710\u22129=4.35\u00d710\u22126\\text{Kb} = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{2.3 \\times 10^{-9}} = 4.35 \\times 10^{-6} [OH\u2212]=Kb\u00d7[BrO\u2212]=4.35\u00d710\u22126\u00d70.240\u22481.044\u00d710\u22126\u22481.02\u00d710\u22123[OH^-] = \\sqrt{Kb \\times [BrO^-]} = \\sqrt{4.35 \\times 10^{-6} \\times 0.240} \u2248 \\sqrt{1.044 \\times 10^{-6}} \u2248 1.02 \\times 10^{-3} pOH=\u2212log\u2061(1.02\u00d710\u22123)\u22482.99\u21d2pH=14\u22122.99=10.34\\text{pOH} = -\\log(1.02 \\times 10^{-3}) \u2248 2.99 \\quad \\Rightarrow \\quad \\text{pH} = 14 – 2.99 = \\boxed{10.34}<\/p>\n\n\n\n
\n\n\n\n\u2705 Final Answers:<\/h3>\n\n\n\n
- \n
- pH after 10.0 mL NaOH added<\/strong>: 8.66<\/strong><\/li>\n\n\n\n
- pH at equivalence point<\/strong>: 10.34<\/strong><\/li>\n<\/ul>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"20.0 ml of 0.200M hypobromous acid, HBrO, is titrated with 0.250M sodium hydroxide, NaOH. Calculate the pH of the solution after the addition of 10.0 ml of NaOH solution. Then Determine 25.0 ml of 0.600M hypobromous acid, HBrO, is titrated with 0.400M sodium hydroxide, NaOH. Calculate the pH of the solution at the equivalence point […]<\/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-219472","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219472","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=219472"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219472\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219472"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219472"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219472"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- pH at equivalence point<\/strong>: 10.34<\/strong><\/li>\n<\/ul>\n\n\n\n
- pH after 10.0 mL NaOH added<\/strong>: 8.66<\/strong><\/li>\n\n\n\n