a) Correct Chemical Equation<\/strong><\/h3>\n\n\n\nThis is an acid-base neutralization reaction between formic acid (HCOOH), a weak acid, and potassium hydroxide (KOH), a strong base:<\/p>\n\n\n\n
HCOOH (aq) + OH\u207b (aq) \u2192 HCOO\u207b (aq) + H\u2082O (l)<\/strong><\/p>\n\n\n\nKOH completely dissociates in water, so the hydroxide ion (OH\u207b) is the actual reacting species.<\/p>\n\n\n\n
\n\n\n\nb) Balanced Chemical Equation & Stoichiometric Relationship<\/strong><\/h3>\n\n\n\nHCOOH (aq) + OH\u207b (aq) \u2192 HCOO\u207b (aq) + H\u2082O (l)<\/strong><\/p>\n\n\n\nStoichiometric relationship:<\/strong> 1 mole of formic acid reacts with 1 mole of hydroxide ion to produce 1 mole of formate ion and 1 mole of water.<\/p>\n\n\n\n
\n\n\n\npH Calculation \u2013 Step-by-Step Explanation<\/h3>\n\n\n\nStep 1: Moles of HCOOH and OH\u207b<\/strong><\/h4>\n\n\n\n\n- Volume HCOOH = 25.0 mL = 0.0250 L<\/li>\n\n\n\n
- Molarity HCOOH = 0.21 M
\u21d2 Moles HCOOH = 0.21 \u00d7 0.0250 = 0.00525 mol<\/strong><\/li>\n\n\n\n- Volume KOH added = 10.0 mL = 0.0100 L<\/li>\n\n\n\n
- Molarity KOH = 0.15 M
\u21d2 Moles OH\u207b = 0.15 \u00d7 0.0100 = 0.00150 mol<\/strong><\/li>\n<\/ul>\n\n\n\nStep 2: Limiting Reactant and Reaction Progress<\/strong><\/h4>\n\n\n\nSince OH\u207b is less than HCOOH, it\u2019s the limiting reactant. It reacts completely with an equal amount of HCOOH:<\/p>\n\n\n\n
\n- Moles of HCOOH remaining = 0.00525 \u2013 0.00150 = 0.00375 mol<\/strong><\/li>\n\n\n\n
- Moles of HCOO\u207b produced = 0.00150 mol<\/li>\n<\/ul>\n\n\n\n
Step 3: Use Henderson-Hasselbalch Equation<\/strong><\/h4>\n\n\n\nWe now have a buffer (HCOOH\/HCOO\u207b). Use the Henderson-Hasselbalch equation<\/strong>:<\/p>\n\n\n\n[
\\text{pH} = \\text{p}K_a + \\log \\left(\\frac{[\\text{A}^-]}{[\\text{HA}]}\\right)
]<\/p>\n\n\n\n
\n- Ka = 1.8 \u00d7 10\u207b\u2074 \u21d2 pKa = \u2013log(1.8 \u00d7 10\u207b\u2074) \u2248 3.74<\/strong><\/li>\n\n\n\n
- Concentrations (in total volume 25.0 + 10.0 = 35.0 mL = 0.035 L):<\/li>\n\n\n\n
- [HCOOH] = 0.00375 \/ 0.035 \u2248 0.107 M<\/strong><\/li>\n\n\n\n
- [HCOO\u207b] = 0.00150 \/ 0.035 \u2248 0.0429 M<\/strong><\/li>\n<\/ul>\n\n\n\n
Now plug into the equation:<\/p>\n\n\n\n
[
\\text{pH} = 3.74 + \\log\\left(\\frac{0.0429}{0.107}\\right) = 3.74 + \\log(0.401) \\approx 3.74 – 0.396 = \\boxed{3.34}
]<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n\n- pH \u2248 3.34<\/strong><\/li>\n\n\n\n
- At this point, the solution is a buffer containing unreacted formic acid and its conjugate base, formate ion. The buffer resists pH change due to the presence of both acidic and basic components in a defined ratio, calculated via Henderson-Hasselbalch.<\/li>\n<\/ul>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/04\/image-198.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"25.0 mL of a 0.21 M formic acid, HCOOH (Ka = 1.8 x 10-4) is titrated with a 0.15M KOH solution. Calculate the pH when 10.0 mL of KOH has been added to the formic acid solution. a) Write the correct equation b) Write the balanced equation and show the stoichiometric relationship in the space […]<\/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-207262","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207262","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=207262"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207262\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207262"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207262"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207262"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
KOH completely dissociates in water, so the hydroxide ion (OH\u207b) is the actual reacting species.<\/p>\n\n\n\n
\n\n\n\n
b) Balanced Chemical Equation & Stoichiometric Relationship<\/strong><\/h3>\n\n\n\nHCOOH (aq) + OH\u207b (aq) \u2192 HCOO\u207b (aq) + H\u2082O (l)<\/strong><\/p>\n\n\n\nStoichiometric relationship:<\/strong> 1 mole of formic acid reacts with 1 mole of hydroxide ion to produce 1 mole of formate ion and 1 mole of water.<\/p>\n\n\n\n
\n\n\n\npH Calculation \u2013 Step-by-Step Explanation<\/h3>\n\n\n\nStep 1: Moles of HCOOH and OH\u207b<\/strong><\/h4>\n\n\n\n\n- Volume HCOOH = 25.0 mL = 0.0250 L<\/li>\n\n\n\n
- Molarity HCOOH = 0.21 M
\u21d2 Moles HCOOH = 0.21 \u00d7 0.0250 = 0.00525 mol<\/strong><\/li>\n\n\n\n- Volume KOH added = 10.0 mL = 0.0100 L<\/li>\n\n\n\n
- Molarity KOH = 0.15 M
\u21d2 Moles OH\u207b = 0.15 \u00d7 0.0100 = 0.00150 mol<\/strong><\/li>\n<\/ul>\n\n\n\nStep 2: Limiting Reactant and Reaction Progress<\/strong><\/h4>\n\n\n\nSince OH\u207b is less than HCOOH, it\u2019s the limiting reactant. It reacts completely with an equal amount of HCOOH:<\/p>\n\n\n\n
\n- Moles of HCOOH remaining = 0.00525 \u2013 0.00150 = 0.00375 mol<\/strong><\/li>\n\n\n\n
- Moles of HCOO\u207b produced = 0.00150 mol<\/li>\n<\/ul>\n\n\n\n
Step 3: Use Henderson-Hasselbalch Equation<\/strong><\/h4>\n\n\n\nWe now have a buffer (HCOOH\/HCOO\u207b). Use the Henderson-Hasselbalch equation<\/strong>:<\/p>\n\n\n\n[
\\text{pH} = \\text{p}K_a + \\log \\left(\\frac{[\\text{A}^-]}{[\\text{HA}]}\\right)
]<\/p>\n\n\n\n
\n- Ka = 1.8 \u00d7 10\u207b\u2074 \u21d2 pKa = \u2013log(1.8 \u00d7 10\u207b\u2074) \u2248 3.74<\/strong><\/li>\n\n\n\n
- Concentrations (in total volume 25.0 + 10.0 = 35.0 mL = 0.035 L):<\/li>\n\n\n\n
- [HCOOH] = 0.00375 \/ 0.035 \u2248 0.107 M<\/strong><\/li>\n\n\n\n
- [HCOO\u207b] = 0.00150 \/ 0.035 \u2248 0.0429 M<\/strong><\/li>\n<\/ul>\n\n\n\n
Now plug into the equation:<\/p>\n\n\n\n
[
\\text{pH} = 3.74 + \\log\\left(\\frac{0.0429}{0.107}\\right) = 3.74 + \\log(0.401) \\approx 3.74 – 0.396 = \\boxed{3.34}
]<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n\n- pH \u2248 3.34<\/strong><\/li>\n\n\n\n
- At this point, the solution is a buffer containing unreacted formic acid and its conjugate base, formate ion. The buffer resists pH change due to the presence of both acidic and basic components in a defined ratio, calculated via Henderson-Hasselbalch.<\/li>\n<\/ul>\n\n\n\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"25.0 mL of a 0.21 M formic acid, HCOOH (Ka = 1.8 x 10-4) is titrated with a 0.15M KOH solution. Calculate the pH when 10.0 mL of KOH has been added to the formic acid solution. a) Write the correct equation b) Write the balanced equation and show the stoichiometric relationship in the space […]<\/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-207262","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207262","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=207262"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207262\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207262"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207262"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207262"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
Stoichiometric relationship:<\/strong> 1 mole of formic acid reacts with 1 mole of hydroxide ion to produce 1 mole of formate ion and 1 mole of water.<\/p>\n\n\n\n Since OH\u207b is less than HCOOH, it\u2019s the limiting reactant. It reacts completely with an equal amount of HCOOH:<\/p>\n\n\n\n We now have a buffer (HCOOH\/HCOO\u207b). Use the Henderson-Hasselbalch equation<\/strong>:<\/p>\n\n\n\n [ Now plug into the equation:<\/p>\n\n\n\n [ 25.0 mL of a 0.21 M formic acid, HCOOH (Ka = 1.8 x 10-4) is titrated with a 0.15M KOH solution. Calculate the pH when 10.0 mL of KOH has been added to the formic acid solution. a) Write the correct equation b) Write the balanced equation and show the stoichiometric relationship in the space […]<\/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-207262","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207262","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=207262"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/207262\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=207262"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=207262"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=207262"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\npH Calculation \u2013 Step-by-Step Explanation<\/h3>\n\n\n\n
Step 1: Moles of HCOOH and OH\u207b<\/strong><\/h4>\n\n\n\n
\n
\u21d2 Moles HCOOH = 0.21 \u00d7 0.0250 = 0.00525 mol<\/strong><\/li>\n\n\n\n
\u21d2 Moles OH\u207b = 0.15 \u00d7 0.0100 = 0.00150 mol<\/strong><\/li>\n<\/ul>\n\n\n\nStep 2: Limiting Reactant and Reaction Progress<\/strong><\/h4>\n\n\n\n
\n
Step 3: Use Henderson-Hasselbalch Equation<\/strong><\/h4>\n\n\n\n
\\text{pH} = \\text{p}K_a + \\log \\left(\\frac{[\\text{A}^-]}{[\\text{HA}]}\\right)
]<\/p>\n\n\n\n\n
\\text{pH} = 3.74 + \\log\\left(\\frac{0.0429}{0.107}\\right) = 3.74 + \\log(0.401) \\approx 3.74 – 0.396 = \\boxed{3.34}
]<\/p>\n\n\n\n
\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
\n
<\/figure>\n","protected":false},"excerpt":{"rendered":"