{"id":218130,"date":"2025-05-23T08:58:56","date_gmt":"2025-05-23T08:58:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218130"},"modified":"2025-05-23T08:58:59","modified_gmt":"2025-05-23T08:58:59","slug":"the-solubility-of-zn3po42-in-water-is-5-9-x-10-5-g-l","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/23\/the-solubility-of-zn3po42-in-water-is-5-9-x-10-5-g-l\/","title":{"rendered":"The solubility of Zn3(PO4)2 in water is = 5.9 x 10-5 g\/L"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">The solubility of Zn3(PO4)2 in water is = 5.9 x 10-5 g\/L, with this given information calculate the solubility product constant (Ksp) of this salt. (Molar Mass Zn3(PO4)2 = 386.11g\/mol).<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Let&#8217;s work through the problem step-by-step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Solubility of Zn\u2083(PO\u2084)\u2082 in water = $5.9 \\times 10^{-5}$ g\/L<\/li>\n\n\n\n<li>Molar mass of Zn\u2083(PO\u2084)\u2082 = 386.11 g\/mol<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Goal:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Calculate the solubility product constant, $K_{sp}$, for Zn\u2083(PO\u2084)\u2082.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Convert solubility from g\/L to mol\/L (molarity)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Solubility (S) in mol\/L is:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>S = \\frac{\\text{solubility in g\/L}}{\\text{molar mass in g\/mol}} = \\frac{5.9 \\times 10^{-5}}{386.11} \\approx 1.53 \\times 10^{-7} \\, \\text{mol\/L}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, the molar solubility $S$ is approximately $1.53 \\times 10^{-7}$ mol\/L.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Write the dissociation equation for Zn\u2083(PO\u2084)\u2082<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\text{Zn}_3(\\text{PO}_4)_2 (s) \\leftrightarrow 3 \\text{Zn}^{2+} (aq) + 2 \\text{PO}_4^{3-} (aq)<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Express ion concentrations in terms of $S$<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">From the dissociation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For every 1 mole of Zn\u2083(PO\u2084)\u2082 dissolved:<\/li>\n\n\n\n<li>$3$ moles of Zn\u00b2\u207a are produced.<\/li>\n\n\n\n<li>$2$ moles of PO\u2084\u00b3\u207b are produced.<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Therefore,<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>[\\text{Zn}^{2+}] = 3S = 3 \\times 1.53 \\times 10^{-7} = 4.59 \\times 10^{-7} \\, \\text{mol\/L}<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>[\\text{PO}_4^{3-}] = 2S = 2 \\times 1.53 \\times 10^{-7} = 3.06 \\times 10^{-7} \\, \\text{mol\/L}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Write the expression for $K_{sp}$<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>K_{sp} = [\\text{Zn}^{2+}]^3 \\times [\\text{PO}_4^{3-}]^2<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Substitute the values:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>K_{sp} = (4.59 \\times 10^{-7})^3 \\times (3.06 \\times 10^{-7})^2<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Calculate powers:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>$(4.59 \\times 10^{-7})^3 = 4.59^3 \\times (10^{-7})^3 = 96.59 \\times 10^{-21} = 9.66 \\times 10^{-20}$<\/li>\n\n\n\n<li>$(3.06 \\times 10^{-7})^2 = 3.06^2 \\times (10^{-7})^2 = 9.36 \\times 10^{-14}$<\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">Multiply:<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>K_{sp} = (9.66 \\times 10^{-20}) \\times (9.36 \\times 10^{-14}) = 9.04 \\times 10^{-33}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final answer:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\boxed{<br>K_{sp} \\approx 9.0 \\times 10^{-33}<br>}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The solubility product constant, $K_{sp}$, quantifies the extent to which a sparingly soluble ionic compound dissolves in water. For Zn\u2083(PO\u2084)\u2082, the solubility is given in grams per liter, but to calculate $K_{sp}$, we first need to convert this to molar solubility (mol\/L) using the molar mass.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Once molar solubility $S$ is known, we use the dissociation stoichiometry to find the concentration of each ion in solution. Zn\u2083(PO\u2084)\u2082 dissociates into 3 zinc ions and 2 phosphate ions per formula unit dissolved, so the ion concentrations are multiples of $S$.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The $K_{sp}$ expression incorporates the ion concentrations raised to the power of their coefficients in the balanced dissociation equation. The very small solubility of Zn\u2083(PO\u2084)\u2082 leads to extremely low ion concentrations, so $K_{sp}$ is correspondingly tiny, indicating the salt is very insoluble.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This approach can be generalized to any ionic salt: find molar solubility from mass solubility, relate ion concentrations via dissociation coefficients, then calculate $K_{sp}$ by raising concentrations to their stoichiometric powers and multiplying.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">In this case, the very low $K_{sp}$ of about $9 \\times 10^{-33}$ reflects the near insolubility of Zn\u2083(PO\u2084)\u2082 in water, consistent with phosphate salts\u2019 typical behavior. This knowledge helps predict precipitation and dissolution equilibria in chemical and environmental systems involving zinc and phosphate ions.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-153.png\" alt=\"\" class=\"wp-image-218131\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The solubility of Zn3(PO4)2 in water is = 5.9 x 10-5 g\/L, with this given information calculate the solubility product constant (Ksp) of this salt. (Molar Mass Zn3(PO4)2 = 386.11g\/mol). The Correct Answer and Explanation is: Let&#8217;s work through the problem step-by-step. Given: Goal: Calculate the solubility product constant, $K_{sp}$, for Zn\u2083(PO\u2084)\u2082. Step 1: Convert [&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-218130","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218130","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=218130"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218130\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218130"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218130"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218130"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}