{"id":246924,"date":"2025-07-07T13:53:03","date_gmt":"2025-07-07T13:53:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=246924"},"modified":"2025-07-07T13:53:05","modified_gmt":"2025-07-07T13:53:05","slug":"calculate-the-ph-when-10-0-ml-of-0-150-m-koh-is-mixed-with-20-0-ml-of-0-300-m-hbro-ka-2-5-10-9","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/07\/calculate-the-ph-when-10-0-ml-of-0-150-m-koh-is-mixed-with-20-0-ml-of-0-300-m-hbro-ka-2-5-10-9\/","title":{"rendered":"Calculate the pH when 10.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 * 10-9"},"content":{"rendered":"\n
Calculate the pH when 10.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 * 10-9<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
To calculate the pH of the solution after mixing 10.0 mL of 0.150 M KOH and 20.0 mL of 0.300 M HBrO, we need to understand the neutralization reaction and the subsequent equilibrium in the solution.<\/p>\n\n\n\n
Step 1: Write the neutralization reaction<\/h3>\n\n\n\n
KOH (a strong base) reacts with HBrO (a weak acid) in a 1:1 ratio:KOH(aq)+HBrO(aq)\u2192KBrO(aq)+H2O(l)\\text{KOH} (aq) + \\text{HBrO} (aq) \\rightarrow \\text{KBrO} (aq) + \\text{H2O} (l)KOH(aq)+HBrO(aq)\u2192KBrO(aq)+H2O(l)<\/p>\n\n\n\n
KOH dissociates completely in water, while HBrO only partially dissociates due to its weak acidic nature.<\/p>\n\n\n\n
Step 2: Calculate the moles of KOH and HBrO<\/h3>\n\n\n\n
\n
Moles of KOH<\/strong>: The concentration of KOH is 0.150 M, and the volume is 10.0 mL = 0.010 L.<\/li>\n<\/ul>\n\n\n\n
moles of KOH=0.150\u2009M\u00d70.010\u2009L=0.0015\u2009moles\\text{moles of KOH} = 0.150 \\, \\text{M} \\times 0.010 \\, \\text{L} = 0.0015 \\, \\text{moles}moles of KOH=0.150M\u00d70.010L=0.0015moles<\/p>\n\n\n\n
\n
Moles of HBrO<\/strong>: The concentration of HBrO is 0.300 M, and the volume is 20.0 mL = 0.020 L.<\/li>\n<\/ul>\n\n\n\n
moles of HBrO=0.300\u2009M\u00d70.020\u2009L=0.0060\u2009moles\\text{moles of HBrO} = 0.300 \\, \\text{M} \\times 0.020 \\, \\text{L} = 0.0060 \\, \\text{moles}moles of HBrO=0.300M\u00d70.020L=0.0060moles<\/p>\n\n\n\n
Step 3: Determine the limiting reagent and excess<\/h3>\n\n\n\n
Since KOH reacts in a 1:1 ratio with HBrO, we can see that KOH is the limiting reagent (0.0015 moles of KOH vs 0.0060 moles of HBrO). After neutralization, we will have:<\/p>\n\n\n\n
\n
0.0015 moles of KOH reacting with 0.0015 moles of HBrO, leaving 0.0045 moles of HBrO unreacted.<\/li>\n<\/ul>\n\n\n\n
Step 4: Calculate the final concentration of HBrO<\/h3>\n\n\n\n
The total volume of the mixture is:Total volume=10.0\u2009mL+20.0\u2009mL=30.0\u2009mL=0.030\u2009L\\text{Total volume} = 10.0 \\, \\text{mL} + 20.0 \\, \\text{mL} = 30.0 \\, \\text{mL} = 0.030 \\, \\text{L}Total volume=10.0mL+20.0mL=30.0mL=0.030L<\/p>\n\n\n\n
The concentration of remaining HBrO is:[HBrO]=0.0045\u2009moles0.030\u2009L=0.150\u2009M[\\text{HBrO}] = \\frac{0.0045 \\, \\text{moles}}{0.030 \\, \\text{L}} = 0.150 \\, \\text{M}[HBrO]=0.030L0.0045moles\u200b=0.150M<\/p>\n\n\n\n
Step 5: Set up the equilibrium expression for HBrO dissociation<\/h3>\n\n\n\n
Since HBrO is a weak acid, it will partially dissociate:HBrO\u21ccH++BrO\u2212\\text{HBrO} \\rightleftharpoons \\text{H}^+ + \\text{BrO}^-HBrO\u21ccH++BrO\u2212<\/p>\n\n\n\n
The equilibrium constant KaK_aKa\u200b for HBrO is given as 2.5\u00d710\u221292.5 \\times 10^{-9}2.5\u00d710\u22129.<\/p>\n\n\n\n
We use the ICE (Initial, Change, Equilibrium) method to calculate the concentration of H+H^+H+ ions.<\/p>\n\n\n\n
[HBrO] decreases by x,[H+] increases by x,[BrO\u2212] increases by x[\\text{HBrO}] \\text{ decreases by } x, \\quad [\\text{H}^+] \\text{ increases by } x, \\quad [\\text{BrO}^-] \\text{ increases by } x[HBrO] decreases by x,[H+] increases by x,[BrO\u2212] increases by x<\/p>\n\n\n\n
Step 6: Write the equilibrium expression<\/h3>\n\n\n\n
The expression for the dissociation of HBrO is:Ka=[H+][BrO\u2212][HBrO]=x\u22c5x0.150\u2212x=2.5\u00d710\u22129K_a = \\frac{[\\text{H}^+][\\text{BrO}^-]}{[\\text{HBrO}]} = \\frac{x \\cdot x}{0.150 – x} = 2.5 \\times 10^{-9}Ka\u200b=[HBrO][H+][BrO\u2212]\u200b=0.150\u2212xx\u22c5x\u200b=2.5\u00d710\u22129<\/p>\n\n\n\n
Simplifying the equation:x20.150\u2212x=2.5\u00d710\u22129\\frac{x^2}{0.150 – x} = 2.5 \\times 10^{-9}0.150\u2212xx2\u200b=2.5\u00d710\u22129<\/p>\n\n\n\n
Since xxx will be very small compared to 0.150 M, we can approximate 0.150\u2212x\u22480.1500.150 – x \\approx 0.1500.150\u2212x\u22480.150, so the equation becomes:x20.150=2.5\u00d710\u22129\\frac{x^2}{0.150} = 2.5 \\times 10^{-9}0.150×2\u200b=2.5\u00d710\u22129<\/p>\n\n\n\n
Thus, the concentration of H+H^+H+ is 6.12\u00d710\u22125\u2009M6.12 \\times 10^{-5} \\, \\text{M}6.12\u00d710\u22125M.<\/p>\n\n\n\n
Step 7: Calculate the pH<\/h3>\n\n\n\n
Finally, we calculate the pH using:pH=\u2212log\u2061[H+]=\u2212log\u2061(6.12\u00d710\u22125)=4.21\\text{pH} = -\\log[\\text{H}^+] = -\\log(6.12 \\times 10^{-5}) = 4.21pH=\u2212log[H+]=\u2212log(6.12\u00d710\u22125)=4.21<\/p>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
The pH of the solution is 4.21<\/strong>.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Calculate the pH when 10.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 * 10-9 The Correct Answer and Explanation is: To calculate the pH of the solution after mixing 10.0 mL of 0.150 M KOH and 20.0 mL of 0.300 M HBrO, we need to […]<\/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-246924","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246924","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=246924"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/246924\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=246924"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=246924"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=246924"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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