{"id":191285,"date":"2025-02-14T11:41:48","date_gmt":"2025-02-14T11:41:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=191285"},"modified":"2025-02-14T11:41:50","modified_gmt":"2025-02-14T11:41:50","slug":"calculate-the-ph-when-20-0-ml-of-0-150-m-koh","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/14\/calculate-the-ph-when-20-0-ml-of-0-150-m-koh\/","title":{"rendered":"Calculate the pH when 20.0 mL of 0.150 M KOH"},"content":{"rendered":"\n<p>Calculate the pH when 20.0 mL of 0.150 M KOH is mixed with 38.0 mL of 0.300 M HBrO (Ka = 2.5 \u00c3\u0192\u00e2\u20ac\u201d 10\u00c3\u00a2\u00c2\ufffd\u00c2\u00bb\u00c3\u00a2\u00c2\ufffd\u00c2\u00b9)<\/p>\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>To calculate the pH of the solution when 20.0 mL of 0.150 M KOH is mixed with 38.0 mL of 0.300 M HBrO, we need to follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Determine the moles of KOH and HBrO<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of KOH:<br>[<br>\\text{Moles of KOH} = Molarity \\times Volume = 0.150 \\, \\text{M} \\times 0.020 \\, \\text{L} = 0.0030 \\, \\text{mol}<br>]<\/li>\n\n\n\n<li>Moles of HBrO:<br>[<br>\\text{Moles of HBrO} = Molarity \\times Volume = 0.300 \\, \\text{M} \\times 0.038 \\, \\text{L} = 0.0114 \\, \\text{mol}<br>]<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Reaction between KOH and HBrO<\/h3>\n\n\n\n<p>KOH is a strong base, and HBrO is a weak acid. The neutralization reaction between KOH and HBrO is:<\/p>\n\n\n\n<p>[<br>\\text{KOH} + \\text{HBrO} \\rightarrow \\text{KBrO} + \\text{H}_2\\text{O}<br>]<\/p>\n\n\n\n<p>Since there are fewer moles of KOH than HBrO, KOH will neutralize an equivalent amount of HBrO. After the reaction, the remaining HBrO will determine the pH.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Moles of HBrO remaining = (0.0114 &#8211; 0.0030 = 0.0084 \\, \\text{mol})<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Concentration of remaining HBrO<\/h3>\n\n\n\n<p>The total volume of the mixture is:<br>[<br>\\text{Volume of solution} = 20.0 \\, \\text{mL} + 38.0 \\, \\text{mL} = 58.0 \\, \\text{mL} = 0.0580 \\, \\text{L}<br>]<\/p>\n\n\n\n<p>The concentration of remaining HBrO is:<br>[<br>[\\text{HBrO}] = \\frac{0.0084 \\, \\text{mol}}{0.0580 \\, \\text{L}} = 0.1448 \\, \\text{M}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Calculate the pH<\/h3>\n\n\n\n<p>HBrO partially dissociates in water, and the dissociation equilibrium is:<\/p>\n\n\n\n<p>[<br>\\text{HBrO} \\rightleftharpoons \\text{H}^+ + \\text{BrO}^-<br>]<\/p>\n\n\n\n<p>The acid dissociation constant ((K_a)) is given as (2.5 \\times 10^{-9}). We can set up an equilibrium expression:<\/p>\n\n\n\n<p>[<br>K_a = \\frac{[\\text{H}^+][\\text{BrO}^-]}{[\\text{HBrO}]}<br>]<\/p>\n\n\n\n<p>Assume that the concentration of ( \\text{H}^+ ) and ( \\text{BrO}^- ) formed at equilibrium is (x), and the concentration of HBrO at equilibrium is (0.1448 &#8211; x). The equation becomes:<\/p>\n\n\n\n<p>[<br>K_a = \\frac{x^2}{0.1448 &#8211; x}<br>]<\/p>\n\n\n\n<p>Given that (K_a) is very small, we can approximate (0.1448 &#8211; x \\approx 0.1448). Thus, the equation simplifies to:<\/p>\n\n\n\n<p>[<br>2.5 \\times 10^{-9} = \\frac{x^2}{0.1448}<br>]<\/p>\n\n\n\n<p>Solving for (x):<\/p>\n\n\n\n<p>[<br>x^2 = (2.5 \\times 10^{-9})(0.1448) = 3.62 \\times 10^{-10}<br>]<br>[<br>x = \\sqrt{3.62 \\times 10^{-10}} = 6.02 \\times 10^{-5} \\, \\text{M}<br>]<\/p>\n\n\n\n<p>Thus, the concentration of ( \\text{H}^+ ) is (6.02 \\times 10^{-5} \\, \\text{M}). To find the pH:<\/p>\n\n\n\n<p>[<br>\\text{pH} = -\\log [\\text{H}^+] = -\\log (6.02 \\times 10^{-5}) = 4.22<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<p>The pH of the solution after mixing 20.0 mL of 0.150 M KOH with 38.0 mL of 0.300 M HBrO is <strong>4.22<\/strong>.<\/p>\n\n\n\n<p>Let me now generate a relevant image for this calculation.<\/p>\n\n\n\n<p>Here is an image that visually illustrates the neutralization between KOH and HBrO, showing the dissociation and pH calculation steps. Let me know if you need further clarification!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-529.png\" alt=\"\" class=\"wp-image-191286\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the pH when 20.0 mL of 0.150 M KOH is mixed with 38.0 mL of 0.300 M HBrO (Ka = 2.5 \u00c3\u0192\u00e2\u20ac\u201d 10\u00c3\u00a2\u00c2\ufffd\u00c2\u00bb\u00c3\u00a2\u00c2\ufffd\u00c2\u00b9) The Correct Answer and Explanation is : To calculate the pH of the solution when 20.0 mL of 0.150 M KOH is mixed with 38.0 mL of 0.300 M HBrO, we [&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-191285","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191285","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=191285"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191285\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191285"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191285"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191285"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}