{"id":191663,"date":"2025-02-15T09:53:06","date_gmt":"2025-02-15T09:53:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=191663"},"modified":"2025-02-15T09:53:08","modified_gmt":"2025-02-15T09:53:08","slug":"the-following-compounds-are-water-soluble","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/15\/the-following-compounds-are-water-soluble\/","title":{"rendered":"The following compounds are water-soluble"},"content":{"rendered":"\n<p>The following compounds are water-soluble. What ions are produced by each compound in aqueous solution?<br>a. KOH<br>b. K\u2082SO\u2084<br>c. LINO<br>d. (NH4)2SO4<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li>Decide whether each of the following is water-soluble. If soluble, tell what ions are produced.<br>a. Na\u2082CO<br>b. CuSO<br>c. NiS<br>d. BaBr<\/li>\n\n\n\n<li>Classify each of these as an acid or a base. Which are strong and which are weak? What ions are produced when each is dissolved in water?<br>a. KOH<br>b. Mg(OH)<br>c. HCIO<br>d. HBr<br>e. LiOH<br>f. H\u2082SO<\/li>\n\n\n\n<li>Balance the following equations and then write the net ionic equation.<br>a. (NH4)2CO3(aq)+Cu(NO3)2 (aq) \u2192CaCO3(s)+NH4NO3(aq)<br>b. Pb(OH),(s)+HCl(aq)\u2192PbCl\u2082(s)+H\u2082O(1)<br>c. BaCO3(s)+HCl(aq)\u2192BaCl2(aq)+H\u2082O(1)+CO2(g)<br>d. CH,CO\u2082H(aq)+Ni(OH)(s)\u2192Ni(CH,CO\u2082):(aq)+H\u2082O(1)<\/li>\n\n\n\n<li>Balance the following equations, and then write the net ionic equation.<br>a. Zn(s)+HCl(aq)\u2192H\u2082(g)+ZnCl2(aq) b. Mg(OH)(s)+HCl(aq)\u2192\u2192MgCl2(aq)+H\u2082O(1)<br>c. HNO3(aq)+CaCO3(s)\u2192Ca(NO\u2083)\u2082(aq)+H\u2082O(1)+CO2(g)<br>d. (NH4)2S(aq)+FeCl (aq)\u2192NH4Cl(aq)+Fe\u2082S(s)<\/li>\n\n\n\n<li>How many entities are present in each of the following?<\/li>\n<\/ol>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>Let&#8217;s break this down:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>Water-soluble Compounds and Their Ions:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a. KOH<\/strong>: Potassium hydroxide (KOH) is soluble in water. When dissolved, it dissociates into <strong>K\u207a (potassium ions)<\/strong> and <strong>OH\u207b (hydroxide ions)<\/strong>.<\/li>\n\n\n\n<li><strong>b. K\u2082SO\u2084<\/strong>: Potassium sulfate (K\u2082SO\u2084) is also water-soluble. It dissociates into <strong>2K\u207a (potassium ions)<\/strong> and <strong>SO\u2084\u00b2\u207b (sulfate ions)<\/strong>.<\/li>\n\n\n\n<li><strong>c. LiNO\u2083<\/strong>: Lithium nitrate (LiNO\u2083) is soluble in water. It dissociates into <strong>Li\u207a (lithium ions)<\/strong> and <strong>NO\u2083\u207b (nitrate ions)<\/strong>.<\/li>\n\n\n\n<li><strong>d. (NH\u2084)\u2082SO\u2084<\/strong>: Ammonium sulfate ((NH\u2084)\u2082SO\u2084) is soluble. It dissociates into <strong>2NH\u2084\u207a (ammonium ions)<\/strong> and <strong>SO\u2084\u00b2\u207b (sulfate ions)<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Solubility and Ion Production:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a. Na\u2082CO\u2083<\/strong>: Sodium carbonate (Na\u2082CO\u2083) is soluble in water. It dissociates into <strong>2Na\u207a (sodium ions)<\/strong> and <strong>CO\u2083\u00b2\u207b (carbonate ions)<\/strong>.<\/li>\n\n\n\n<li><strong>b. CuSO\u2084<\/strong>: Copper(II) sulfate (CuSO\u2084) is soluble. It dissociates into <strong>Cu\u00b2\u207a (copper ions)<\/strong> and <strong>SO\u2084\u00b2\u207b (sulfate ions)<\/strong>.<\/li>\n\n\n\n<li><strong>c. NiS<\/strong>: Nickel(II) sulfide (NiS) is insoluble in water.<\/li>\n\n\n\n<li><strong>d. BaBr\u2082<\/strong>: Barium bromide (BaBr\u2082) is soluble. It dissociates into <strong>Ba\u00b2\u207a (barium ions)<\/strong> and <strong>2Br\u207b (bromide ions)<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">3. <strong>Acid\/Base Classification and Ion Production:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a. KOH<\/strong>: Potassium hydroxide is a strong base and dissociates completely into <strong>K\u207a<\/strong> and <strong>OH\u207b<\/strong>.<\/li>\n\n\n\n<li><strong>b. Mg(OH)\u2082<\/strong>: Magnesium hydroxide is a weak base. It dissociates partially into <strong>Mg\u00b2\u207a<\/strong> and <strong>OH\u207b<\/strong>.<\/li>\n\n\n\n<li><strong>c. HClO<\/strong>: Hypochlorous acid (HClO) is a weak acid. It dissociates into <strong>H\u207a<\/strong> and <strong>ClO\u207b<\/strong>.<\/li>\n\n\n\n<li><strong>d. HBr<\/strong>: Hydrobromic acid (HBr) is a strong acid and dissociates completely into <strong>H\u207a<\/strong> and <strong>Br\u207b<\/strong>.<\/li>\n\n\n\n<li><strong>e. LiOH<\/strong>: Lithium hydroxide is a strong base and dissociates completely into <strong>Li\u207a<\/strong> and <strong>OH\u207b<\/strong>.<\/li>\n\n\n\n<li><strong>f. H\u2082SO\u2084<\/strong>: Sulfuric acid (H\u2082SO\u2084) is a strong acid in the first dissociation step (H\u2082SO\u2084 \u2192 H\u207a + HSO\u2084\u207b) but weak in the second dissociation (HSO\u2084\u207b \u21cc H\u207a + SO\u2084\u00b2\u207b).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">4. <strong>Balancing Chemical Equations:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a.<\/strong><br>Balanced equation:<br>[(NH\u2084)\u2082CO\u2083(aq) + Cu(NO\u2083)\u2082(aq) \u2192 CuCO\u2083(s) + 2NH\u2084NO\u2083(aq)]<br>Net ionic equation:<br>[CO\u2083\u00b2\u207b + Cu\u00b2\u207a \u2192 CuCO\u2083(s)]<\/li>\n\n\n\n<li><strong>b.<\/strong><br>Balanced equation:<br>[Pb(OH)\u2082(s) + 2HCl(aq) \u2192 PbCl\u2082(s) + 2H\u2082O(l)]<br>Net ionic equation:<br>[Pb(OH)\u2082(s) + 2H\u207a(aq) \u2192 Pb\u00b2\u207a(aq) + 2H\u2082O(l)]<\/li>\n\n\n\n<li><strong>c.<\/strong><br>Balanced equation:<br>[BaCO\u2083(s) + 2HCl(aq) \u2192 BaCl\u2082(aq) + H\u2082O(l) + CO\u2082(g)]<br>Net ionic equation:<br>[BaCO\u2083(s) + 2H\u207a(aq) \u2192 Ba\u00b2\u207a(aq) + CO\u2082(g) + H\u2082O(l)]<\/li>\n\n\n\n<li><strong>d.<\/strong><br>Balanced equation:<br>[CH\u2083COOH(aq) + Ni(OH)\u2082(s) \u2192 Ni(CH\u2083COO)\u2082(aq) + H\u2082O(l)]<br>Net ionic equation:<br>[CH\u2083COOH(aq) + Ni(OH)\u2082(s) \u2192 Ni\u00b2\u207a(aq) + 2CH\u2083COO\u207b(aq) + 2H\u2082O(l)]<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">5. <strong>Additional Balanced Equations:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>a.<\/strong><br>Balanced equation:<br>[Zn(s) + 2HCl(aq) \u2192 H\u2082(g) + ZnCl\u2082(aq)]<br>Net ionic equation:<br>[Zn(s) + 2H\u207a(aq) \u2192 Zn\u00b2\u207a(aq) + H\u2082(g)]<\/li>\n\n\n\n<li><strong>b.<\/strong><br>Balanced equation:<br>[Mg(OH)\u2082(s) + 2HCl(aq) \u2192 MgCl\u2082(aq) + 2H\u2082O(l)]<br>Net ionic equation:<br>[Mg(OH)\u2082(s) + 2H\u207a(aq) \u2192 Mg\u00b2\u207a(aq) + 2H\u2082O(l)]<\/li>\n\n\n\n<li><strong>c.<\/strong><br>Balanced equation:<br>[2HNO\u2083(aq) + CaCO\u2083(s) \u2192 Ca(NO\u2083)\u2082(aq) + H\u2082O(l) + CO\u2082(g)]<br>Net ionic equation:<br>[2H\u207a(aq) + CO\u2083\u00b2\u207b(s) \u2192 H\u2082O(l) + CO\u2082(g)]<\/li>\n\n\n\n<li><strong>d.<\/strong><br>Balanced equation:<br>[(NH\u2084)\u2082S(aq) + FeCl\u2082(aq) \u2192 2NH\u2084Cl(aq) + FeS(s)]<br>Net ionic equation:<br>[S\u00b2\u207b(aq) + Fe\u00b2\u207a(aq) \u2192 FeS(s)]<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">6. <strong>Entities in Solutions:<\/strong><\/h3>\n\n\n\n<p>Let me provide an explanation for counting entities in solution:<\/p>\n\n\n\n<p>Each compound in water dissociates into ions. To count how many entities are present in solution, consider the number of molecules (or formula units) of the compound that dissociate and the ions that they form.<\/p>\n\n\n\n<p>For example:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>KOH<\/strong> dissociates into <strong>K\u207a<\/strong> and <strong>OH\u207b<\/strong>, so for 1 formula unit of KOH, you get 2 ions.<\/li>\n\n\n\n<li><strong>Na\u2082CO\u2083<\/strong> dissociates into <strong>2Na\u207a<\/strong> and <strong>CO\u2083\u00b2\u207b<\/strong>, so 1 formula unit gives 3 entities (2 Na\u207a + 1 CO\u2083\u00b2\u207b).<\/li>\n<\/ul>\n\n\n\n<p>This counting works similarly for other compounds.<\/p>\n\n\n\n<p>Let me generate the visual of this!<\/p>\n\n\n\n<p>Here\u2019s a visual representation of the number of entities produced when each compound dissolves in water. The entities are counted based on how the compound dissociates into ions. For example, KOH produces 2 entities (K\u207a and OH\u207b), while Na\u2082CO\u2083 produces 3 (2 Na\u207a and CO\u2083\u00b2\u207b). If you have more questions or need further clarifications, feel free to ask!<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-622-1024x813.png\" alt=\"\" class=\"wp-image-191664\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The following compounds are water-soluble. What ions are produced by each compound in aqueous solution?a. KOHb. K\u2082SO\u2084c. LINOd. (NH4)2SO4 The Correct Answer and Explanation is : Let&#8217;s break this down: 1. Water-soluble Compounds and Their Ions: 2. Solubility and Ion Production: 3. Acid\/Base Classification and Ion Production: 4. Balancing Chemical Equations: 5. Additional Balanced Equations: [&hellip;]<\/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-191663","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191663","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=191663"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191663\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191663"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191663"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191663"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}