A 0.56-g sample of barium hydroxide, Ba(OH)2 (molar mass 171.35 g\/mol), is dissolved in enough water to make 250.0 mL of solution. What is the pH of the resulting solution?<\/p>\n\n\n\n
(A) pH = 1.58 (B) pH- 1.88 (C) pH 12.12 (D) pH 12.42<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the pH of the resulting solution, we need to consider the dissociation of barium hydroxide (Ba(OH)\u2082) in water. Ba(OH)\u2082 is a strong base, so it dissociates completely in water into barium ions (Ba\u00b2\u207a) and hydroxide ions (OH\u207b).<\/p>\n\n\n\n The dissociation reaction is:<\/p>\n\n\n\n [ We are given that the sample of Ba(OH)\u2082 weighs 0.56 g, and the molar mass of Ba(OH)\u2082 is 171.35 g\/mol. To find the moles of Ba(OH)\u2082:<\/p>\n\n\n\n [ [ Since Ba(OH)\u2082 dissociates completely, for every mole of Ba(OH)\u2082, we get 2 moles of OH\u207b. Thus, the number of moles of OH\u207b is:<\/p>\n\n\n\n [ Now, we calculate the concentration of OH\u207b in the solution. The volume of the solution is 250.0 mL (0.2500 L):<\/p>\n\n\n\n [ The pOH is calculated using the formula:<\/p>\n\n\n\n [ [ Finally, the pH can be found using the relationship between pH and pOH:<\/p>\n\n\n\n [ [ The pH of the solution is 12.42<\/strong>, so the correct answer is (D) pH = 12.42<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" A 0.56-g sample of barium hydroxide, Ba(OH)2 (molar mass 171.35 g\/mol), is dissolved in enough water to make 250.0 mL of solution. What is the pH of the resulting solution? (A) pH = 1.58 (B) pH- 1.88 (C) pH 12.12 (D) pH 12.42 The correct answer and explanation is : To calculate the pH of […]<\/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-197700","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197700","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=197700"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197700\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197700"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197700"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197700"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\\text{Ba(OH)}_2 (s) \\rightarrow \\text{Ba}^{2+} (aq) + 2 \\text{OH}^- (aq)
]<\/p>\n\n\n\nStep 1: Calculate the moles of Ba(OH)\u2082<\/h3>\n\n\n\n
\\text{moles of Ba(OH)}_2 = \\frac{\\text{mass of Ba(OH)}_2}{\\text{molar mass of Ba(OH)}_2}
]<\/p>\n\n\n\n
\\text{moles of Ba(OH)}_2 = \\frac{0.56 \\, \\text{g}}{171.35 \\, \\text{g\/mol}} = 0.00327 \\, \\text{mol}
]<\/p>\n\n\n\nStep 2: Determine the concentration of OH\u207b<\/h3>\n\n\n\n
\\text{moles of OH}^- = 2 \\times \\text{moles of Ba(OH)}_2 = 2 \\times 0.00327 \\, \\text{mol} = 0.00654 \\, \\text{mol}
]<\/p>\n\n\n\n
\\text{[OH}^-] = \\frac{\\text{moles of OH}^-}{\\text{volume of solution in L}} = \\frac{0.00654 \\, \\text{mol}}{0.2500 \\, \\text{L}} = 0.02616 \\, \\text{M}
]<\/p>\n\n\n\nStep 3: Calculate the pOH<\/h3>\n\n\n\n
\\text{pOH} = -\\log[\\text{OH}^-]
]<\/p>\n\n\n\n
\\text{pOH} = -\\log(0.02616) = 1.58
]<\/p>\n\n\n\nStep 4: Calculate the pH<\/h3>\n\n\n\n
\\text{pH} + \\text{pOH} = 14
]<\/p>\n\n\n\n
\\text{pH} = 14 – \\text{pOH} = 14 – 1.58 = 12.42
]<\/p>\n\n\n\nConclusion:<\/h3>\n\n\n\n