The key reason Joe and Jane’s titrations required the same volume of EDTA solution, despite different concentrations and compositions of metal ions in their samples, is because EDTA is a chelating agent that binds with both Ca\u00b2\u207a and Mg\u00b2\u207a ions at a 1:1 molar ratio. Here\u2019s a breakdown of the situation:<\/p>\n\n\n\n
- \n
- Joe’s Sample<\/strong>: 100 mL of a solution containing 1.0 mM Ca\u00b2\u207a.<\/li>\n<\/ol>\n\n\n\n
- \n
- Moles of Ca\u00b2\u207a in Joe\u2019s sample = (1.0 \\, \\text{mM} \\times 0.100 \\, \\text{L} = 0.100 \\, \\text{mmol}).<\/li>\n<\/ul>\n\n\n\n
- \n
- Jane’s Sample<\/strong>: 100 mL of a solution containing 0.50 mM Ca\u00b2\u207a and 0.50 mM Mg\u00b2\u207a.<\/li>\n<\/ol>\n\n\n\n
- \n
- Total moles of cations in Jane\u2019s sample = (0.50 \\, \\text{mM} \\times 0.100 \\, \\text{L} + 0.50 \\, \\text{mM} \\times 0.100 \\, \\text{L} = 0.050 \\, \\text{mmol Ca\u00b2\u207a} + 0.050 \\, \\text{mmol Mg\u00b2\u207a} = 0.100 \\, \\text{mmol}).<\/li>\n<\/ul>\n\n\n\n
Since both samples have the same total concentration of divalent metal ions that can complex with EDTA, they require the same amount of EDTA to reach the endpoint. EDTA complexes with both calcium and magnesium ions in a stoichiometric 1:1 ratio. This stoichiometry is crucial in titrations because it means that each mole of EDTA will bind one mole of either Ca\u00b2\u207a or Mg\u00b2\u207a. Therefore, the total number of moles of EDTA needed to titrate each sample is equal to the total moles of metal ions present in each sample, regardless of the type of metal ion.<\/p>\n\n\n\n
This outcome emphasizes the versatility and utility of EDTA in complexometric titrations, where it is used to determine concentrations of a variety of metal ions by forming stable complexes. The ability of EDTA to bind equally well with different metal ions under similar conditions makes it a powerful tool in analytical chemistry for the quantification of metal ions in a mixture.<\/p>\n\n\n\n
Now, I’ll generate the image for you.<\/p>\n\n\n\n
Here is the image of the chemistry laboratory scene with Joe and Jane performing their titrations. It visually represents how both chemists are using EDTA to titrate their respective samples under laboratory conditions.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1150.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Joe the chemist titrated 100 mL of a sample that was 1.0 mM in Ca2+ with a standard EDTA solution to the EBT endpoint. Jane the chemist titrated a 100 mL sample that was 0.50 mM in Ca2+ and 0.50 mM in Mg2+ with the same EDTA standard solution. Both required exactly the same volume […]<\/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-193855","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193855","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=193855"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193855\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193855"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193855"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193855"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Total moles of cations in Jane\u2019s sample = (0.50 \\, \\text{mM} \\times 0.100 \\, \\text{L} + 0.50 \\, \\text{mM} \\times 0.100 \\, \\text{L} = 0.050 \\, \\text{mmol Ca\u00b2\u207a} + 0.050 \\, \\text{mmol Mg\u00b2\u207a} = 0.100 \\, \\text{mmol}).<\/li>\n<\/ul>\n\n\n\n
- Jane’s Sample<\/strong>: 100 mL of a solution containing 0.50 mM Ca\u00b2\u207a and 0.50 mM Mg\u00b2\u207a.<\/li>\n<\/ol>\n\n\n\n
- Moles of Ca\u00b2\u207a in Joe\u2019s sample = (1.0 \\, \\text{mM} \\times 0.100 \\, \\text{L} = 0.100 \\, \\text{mmol}).<\/li>\n<\/ul>\n\n\n\n