{"id":186062,"date":"2025-01-24T05:36:17","date_gmt":"2025-01-24T05:36:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186062"},"modified":"2025-01-24T05:36:19","modified_gmt":"2025-01-24T05:36:19","slug":"calculate-oh-and-ph-for","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/24\/calculate-oh-and-ph-for\/","title":{"rendered":"Calculate [OH – ] and pH for"},"content":{"rendered":"\n
    \n
  1. Calculate [OH – ] and pH for<\/li>\n<\/ol>\n\n\n\n

    (a) 1.5 X 10 -3 M Sr(OH) 2<\/p>\n\n\n\n

    (b) 2.250g of LiOH in 250.0 mL of solution<\/p>\n\n\n\n

    (c) 1.00 mL of 0.175 M NaOH diluted to 2.00 L<\/p>\n\n\n\n

    (d) a soution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 X 10 -2 M Ca(OH) 2<\/p>\n\n\n\n

    Group of answer choices
    A.) a) [OH – ]= 3.0 x 10 -3 M, pH= 11.48<\/p>\n\n\n\n

    b) [OH – ]= 0.3758 M, pH= 13.5750<\/p>\n\n\n\n

    c) [OH – ]= 8.75 x 10 -5 M, pH= 4.06<\/p>\n\n\n\n

    d) [OH – ]= 0.17 M, pH= 13.23<\/p>\n\n\n\n

    B.) a) [OH – ]= 3.0 x 10 -3 M, pH= 11.48<\/p>\n\n\n\n

    b) [OH – ]= 0.3758 M, pH= 13.5750<\/p>\n\n\n\n

    c) [OH – ]= 8.75 x 10 -5 M, pH= 9.942<\/p>\n\n\n\n

    d) [OH – ]= 0.17 M, pH= 13.23<\/p>\n\n\n\n

    C.) a) [OH – ]= 1.5 x 10 -3 M, pH= 11.18<\/p>\n\n\n\n

    b) [OH – ]= 0.3758 M, pH= 13.5750<\/p>\n\n\n\n

    c) [OH – ]= 8.75 x 10 -5 M, pH= 9.942<\/p>\n\n\n\n

    d) [OH – ]= 0.17 M, pH= 13.23<\/p>\n\n\n\n

    D.) a) [OH – ]= 3.0 x 10 -3 M, pH= 2.52<\/p>\n\n\n\n

    b) [OH – ]= 0.3758 M, pH= 0.425<\/p>\n\n\n\n

    c) [OH – ]= 8.75 x 10 -5 M, pH= 4.06<\/p>\n\n\n\n

    d) [OH – ]= 0.17 M, pH= 13.23<\/p>\n\n\n\n

    The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

    To determine the hydroxide ion concentration ([OH^-]) and pH for each solution, we will analyze each case individually:<\/p>\n\n\n\n

    (a) 1.5 \u00d7 10\u207b\u00b3 M Sr(OH)\u2082<\/strong><\/p>\n\n\n\n

    Strontium hydroxide (Sr(OH)\u2082) is a strong base that dissociates completely in water:<\/p>\n\n\n\n

    [ \\text{Sr(OH)}_2 \\rightarrow \\text{Sr}^{2+} + 2\\text{OH}^- ]<\/p>\n\n\n\n

    Given the concentration of Sr(OH)\u2082 is 1.5 \u00d7 10\u207b\u00b3 M, the concentration of hydroxide ions ([OH^-]) will be twice this value:<\/p>\n\n\n\n

    [ [OH^-] = 2 \\times 1.5 \\times 10^{-3} \\, \\text{M} = 3.0 \\times 10^{-3} \\, \\text{M} ]<\/p>\n\n\n\n

    The pOH is calculated as:<\/p>\n\n\n\n

    [ \\text{pOH} = -\\log[OH^-] = -\\log(3.0 \\times 10^{-3}) \\approx 2.52 ]<\/p>\n\n\n\n

    Since pH + pOH = 14, the pH is:<\/p>\n\n\n\n

    [ \\text{pH} = 14 – 2.52 = 11.48 ]<\/p>\n\n\n\n

    (b) 2.250 g of LiOH in 250.0 mL of solution<\/strong><\/p>\n\n\n\n

    First, calculate the moles of LiOH:<\/p>\n\n\n\n

    [ \\text{Molar mass of LiOH} = 6.939 + 15.999 + 1.008 = 23.946 \\, \\text{g\/mol} ]<\/p>\n\n\n\n

    [ \\text{Moles of LiOH} = \\frac{2.250 \\, \\text{g}}{23.946 \\, \\text{g\/mol}} \\approx 0.0939 \\, \\text{mol} ]<\/p>\n\n\n\n

    The volume of the solution is 250.0 mL, or 0.2500 L. Therefore, the concentration of LiOH is:<\/p>\n\n\n\n

    [ [\\text{LiOH}] = \\frac{0.0939 \\, \\text{mol}}{0.2500 \\, \\text{L}} = 0.3756 \\, \\text{M} ]<\/p>\n\n\n\n

    Since LiOH is a strong base, it dissociates completely:<\/p>\n\n\n\n

    [ \\text{LiOH} \\rightarrow \\text{Li}^+ + \\text{OH}^- ]<\/p>\n\n\n\n

    Thus, ([OH^-] = 0.3756 \\, \\text{M}). The pOH is:<\/p>\n\n\n\n

    [ \\text{pOH} = -\\log(0.3756) \\approx 0.425 ]<\/p>\n\n\n\n

    And the pH is:<\/p>\n\n\n\n

    [ \\text{pH} = 14 – 0.425 = 13.575 ]<\/p>\n\n\n\n

    (c) 1.00 mL of 0.175 M NaOH diluted to 2.00 L<\/strong><\/p>\n\n\n\n

    The moles of NaOH before dilution are:<\/p>\n\n\n\n

    [ \\text{Moles of NaOH} = 0.175 \\, \\text{M} \\times 0.00100 \\, \\text{L} = 1.75 \\times 10^{-4} \\, \\text{mol} ]<\/p>\n\n\n\n

    After dilution to 2.00 L, the concentration of NaOH is:<\/p>\n\n\n\n

    [ [\\text{NaOH}] = \\frac{1.75 \\times 10^{-4} \\, \\text{mol}}{2.00 \\, \\text{L}} = 8.75 \\times 10^{-5} \\, \\text{M} ]<\/p>\n\n\n\n

    Since NaOH is a strong base, it dissociates completely:<\/p>\n\n\n\n

    [ [OH^-] = 8.75 \\times 10^{-5} \\, \\text{M} ]<\/p>\n\n\n\n

    The pOH is:<\/p>\n\n\n\n

    [ \\text{pOH} = -\\log(8.75 \\times 10^{-5}) \\approx 4.06 ]<\/p>\n\n\n\n

    And the pH is:<\/p>\n\n\n\n

    [ \\text{pH} = 14 – 4.06 = 9.94 ]<\/p>\n\n\n\n

    (d) A solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 \u00d7 10\u207b\u00b2 M Ca(OH)\u2082<\/strong><\/p>\n\n\n\n

    First, calculate the moles of KOH:<\/p>\n\n\n\n

    [ \\text{Moles of KOH} = 0.105 \\, \\text{M} \\times 0.00500 \\, \\text{L} = 5.25 \\times 10^{-4} \\, \\text{mol} ]<\/p>\n\n\n\n

    Next, calculate the moles of Ca(OH)\u2082:<\/p>\n\n\n\n

    [ \\text{Moles of Ca(OH)}_2 = 9.5 \\times 10^{-2} \\, \\text{M} \\times 0.0150 \\, \\text{L} = 1.425 \\times 10^{-3} \\, \\text{mol} ]<\/p>\n\n\n\n

    The total volume of the solution is:<\/p>\n\n\n\n

    [ 5.00 \\, \\text{mL} + 15.0 \\, \\text{mL} = 20.0 \\, \\text{mL} = 0.0200 \\, \\text{L} ]<\/p>\n\n\n\n

    The total moles of hydroxide ions from both sources are:<\/p>\n\n\n\n

    [ 5.25 \\times 10^{-4} \\, \\text{mol} + 1.425 \\times 10^{-3} \\,<\/p>\n","protected":false},"excerpt":{"rendered":"

    (a) 1.5 X 10 -3 M Sr(OH) 2 (b) 2.250g of LiOH in 250.0 mL of solution (c) 1.00 mL of 0.175 M NaOH diluted to 2.00 L (d) a soution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 9.5 X 10 -2 M Ca(OH) 2 Group of answer choicesA.) […]<\/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-186062","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186062","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=186062"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186062\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186062"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186062"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186062"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}