{"id":348,"date":"2025-05-07T19:40:16","date_gmt":"2025-05-07T19:40:16","guid":{"rendered":"https:\/\/yaveni.com\/blog\/?p=348"},"modified":"2025-05-07T19:40:40","modified_gmt":"2025-05-07T19:40:40","slug":"calculate-the-solubility-of-potassium-bromide-at-23-degrees-c","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculate-the-solubility-of-potassium-bromide-at-23-degrees-c\/","title":{"rendered":"calculate the solubility of potassium bromide at 23 degrees C"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">calculate the solubility of potassium bromide at 23 degrees C. HInt: Assume that the solubility increases by an equal amount for each degree between 20 degrees C and 30 degrees C.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To calculate the solubility of potassium bromide (KBr) at <strong>23\u00b0C<\/strong>, we need reference values for the solubility at <strong>20\u00b0C<\/strong> and <strong>30\u00b0C<\/strong>. These values are commonly found in solubility charts.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">From a standard solubility chart:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At <strong>20\u00b0C<\/strong>, the solubility of KBr is approximately <strong>65 g\/100 g of water<\/strong>.<\/li>\n\n\n\n<li>At <strong>30\u00b0C<\/strong>, the solubility of KBr is approximately <strong>74 g\/100 g of water<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Determine the total increase in solubility from 20\u00b0C to 30\u00b0C<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>74\\,g &#8211; 65\\,g = 9\\,g<br>$$<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">So, over a 10\u00b0C interval (from 20\u00b0C to 30\u00b0C), the solubility increases by <strong>9 g<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the increase in solubility per 1\u00b0C<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>\\frac{9\\,g}{10^\\circ C} = 0.9\\,g\/^\\circ C<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Determine the increase from 20\u00b0C to 23\u00b0C (which is 3\u00b0C)<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>0.9\\,g\/^\\circ C \\times 3^\\circ C = 2.7\\,g<br>$$<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Add the increase to the solubility at 20\u00b0C<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">$$<br>65\\,g + 2.7\\,g = 67.7\\,g\/100\\,g\\,\\text{of water}<br>$$<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Final Answer: <strong>67.7 g of KBr per 100 g of water at 23\u00b0C<\/strong><\/h3>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u270f\ufe0f Detailed Explanation (300+ words):<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">Solubility refers to the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature to form a saturated solution. In this case, we&#8217;re calculating how much potassium bromide (KBr), an ionic salt, can dissolve in water at 23\u00b0C.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The solubility of most ionic compounds like KBr increases with temperature. This is because increasing temperature provides more kinetic energy to water molecules, allowing them to break the ionic bonds in the salt more efficiently and surround the dissociated ions (K\u207a and Br\u207b), resulting in increased solubility.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">The question provides a temperature range (20\u00b0C to 30\u00b0C) and asks us to assume the solubility increases linearly across that range. A linear increase means the solubility change is consistent for each degree Celsius. From solubility data, we know that KBr has a solubility of 65 g per 100 g of water at 20\u00b0C and 74 g per 100 g of water at 30\u00b0C. The total increase over 10 degrees is 9 g, so the increase per degree is 0.9 g.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Since 23\u00b0C is 3\u00b0C above 20\u00b0C, we multiply the 0.9 g\/\u00b0C by 3 to get an increase of 2.7 g. Adding this to the base solubility at 20\u00b0C (65 g), we get a final solubility of 67.7 g at 23\u00b0C.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">This method is useful when exact solubility data isn\u2019t available for the target temperature but values are known at two reference points. Linear interpolation is an estimation technique that assumes a straight-line relationship between two known values\u2014a reasonable assumption over small temperature intervals for many salts.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>calculate the solubility of potassium bromide at 23 degrees C. HInt: Assume that the solubility increases by an equal amount for each degree between 20 degrees C and 30 degrees C. The correct answer and explanation is : To calculate the solubility of potassium bromide (KBr) at 23\u00b0C, we need reference values for the solubility [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-348","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/348","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=348"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/348\/revisions"}],"predecessor-version":[{"id":349,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/348\/revisions\/349"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=348"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=348"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=348"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}