To calculate the concentration of sodium ions and carbonate ions in the solution, we need to follow these steps:<\/p>\n\n\n\n
- \n
- Write the Dissociation Equation:<\/strong> Sodium carbonate (Na\u2082CO\u2083) dissociates in water according to the following equation:
[
\\text{Na}_2\\text{CO}_3 (s) \\rightarrow 2\\text{Na}^+ (aq) + \\text{CO}_3^{2-} (aq)
]
This equation tells us that each formula unit of sodium carbonate produces two sodium ions and one carbonate ion.<\/li>\n\n\n\n- Calculate the Molar Mass of Sodium Carbonate:<\/strong> The molar mass of Na\u2082CO\u2083 is calculated by summing the molar masses of its constituent elements.<\/li>\n<\/ol>\n\n\n\n
- \n
- Sodium (Na): (22.99 \\, \\text{g\/mol})<\/li>\n\n\n\n
- Carbon (C): (12.01 \\, \\text{g\/mol})<\/li>\n\n\n\n
- Oxygen (O): (16.00 \\, \\text{g\/mol}) per atom and there are three oxygen atoms.
[
\\text{Molar mass of Na}_2\\text{CO}_3 = 2(22.99) + 12.01 + 3(16.00) = 105.99 \\, \\text{g\/mol}
]<\/li>\n<\/ul>\n\n\n\n- \n
- Calculate Moles of Sodium Carbonate Dissolved:<\/strong>
[
\\text{Moles of Na}_2\\text{CO}_3 = \\frac{8.50 \\, \\text{g}}{105.99 \\, \\text{g\/mol}} \\approx 0.0802 \\, \\text{mol}
]<\/li>\n\n\n\n- Calculate Moles of Ions Produced:<\/strong> From the stoichiometry of the dissociation equation, each mole of Na\u2082CO\u2083 produces two moles of Na\u207a and one mole of CO\u2083\u00b2\u207b.<\/li>\n<\/ol>\n\n\n\n
- \n
- Moles of Na\u207a: (2 \\times 0.0802 \\, \\text{mol} = 0.1604 \\, \\text{mol})<\/li>\n\n\n\n
- Moles of CO\u2083\u00b2\u207b: (0.0802 \\, \\text{mol})<\/li>\n<\/ul>\n\n\n\n
- \n
- Calculate the Concentrations:<\/strong><\/li>\n<\/ol>\n\n\n\n
- \n
- Volume of solution = 500 mL = 0.500 L<\/li>\n\n\n\n
- Concentration of Na\u207a ions:
[
[\\text{Na}^+] = \\frac{0.1604 \\, \\text{mol}}{0.500 \\, \\text{L}} = 0.3208 \\, \\text{M}
]<\/li>\n\n\n\n - Concentration of CO\u2083\u00b2\u207b ions:
[
[\\text{CO}_3^{2-}] = \\frac{0.0802 \\, \\text{mol}}{0.500 \\, \\text{L}} = 0.1604 \\, \\text{M}
]<\/li>\n<\/ul>\n\n\n\nThis calculation demonstrates the process of determining the concentration of ions in a solution from a known quantity of a solid salt. Understanding the stoichiometry of the salt’s dissociation in water is critical. Each mole of sodium carbonate provides specific amounts of sodium and carbonate ions, directly affecting the ionic concentration in the solution.<\/p>\n\n\n\n
For the image, let’s create a visual that illustrates the dissociation of sodium carbonate in water, showing the formula units breaking into ions and their respective concentrations.<\/p>\n\n\n\n
Here is a visual illustration of the dissociation of sodium carbonate (Na\u2082CO\u2083) in water, showing how the compound splits into sodium ions (Na\u207a) and carbonate ions (CO\u2083\u00b2\u207b), along with their calculated concentrations in the solution. This image can help in visualizing the chemical process and understanding the concept better.<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1153.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"Standard solutions of sodium carbonate, Na2CO3(aq), are required for certain types of chemical analyses. If 8.50 g ofsodium carbonate is dissolved in 500 mL of distilled water, calculate the concentration of the sodium ions and the concentration of the carbonate ions dissolved in this solution.\u00c3\u201a\u00c2 Hint: In order to solve this problem, write the correctly […]<\/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-193867","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193867","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=193867"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193867\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193867"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193867"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193867"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Calculate the Concentrations:<\/strong><\/li>\n<\/ol>\n\n\n\n
- Calculate Moles of Ions Produced:<\/strong> From the stoichiometry of the dissociation equation, each mole of Na\u2082CO\u2083 produces two moles of Na\u207a and one mole of CO\u2083\u00b2\u207b.<\/li>\n<\/ol>\n\n\n\n
- Calculate Moles of Sodium Carbonate Dissolved:<\/strong>
- Calculate the Molar Mass of Sodium Carbonate:<\/strong> The molar mass of Na\u2082CO\u2083 is calculated by summing the molar masses of its constituent elements.<\/li>\n<\/ol>\n\n\n\n