{"id":234144,"date":"2025-06-13T19:08:56","date_gmt":"2025-06-13T19:08:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=234144"},"modified":"2025-06-13T19:08:58","modified_gmt":"2025-06-13T19:08:58","slug":"based-on-the-red-ink-corrections-solve","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/based-on-the-red-ink-corrections-solve\/","title":{"rendered":"Based on the red ink corrections, solve"},"content":{"rendered":"\n
Based on the red ink corrections, solve. The maximum solubility of NaHSO4 is 420 g\/L @ 25\u00b0C. Its molar mass is 104.061 g\/mol. The pKa of HSO4^- is 1.92. What is the pH of a saturated solution of NaHSO4? (6 pts.) Plan: 1.92 The maximum solubility of NaHSO4 is 420 g\/L @ 25\u00b0C. Its molar mass is 104.061 g\/mol. The pKa of HSO4^- is 1.92. What is the pH of a saturated solution of NaHSO4? (6 pts.) max: 420 g\/L @ 25\u00b0C Prc: 1.92 420 g\/mol x 1 mol\/104.061 g = 4.033 mol\/L SO4^-2 + H2O -> HSO4^- + OH^- 1.92 = -log([HSO4^-]\/[SO4^-2]) 1.92 = -log(4.033\/x) x = 0.033 M pH = 1\/2(pKa - log([A-]\/[HA])) pH = 1\/2(1.92 - log(0.033\/4.033)) pH = 1\/2(1.92 - 1.39) pH = 0.265<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Based on the provided information, the correct solution is as follows:<\/p>\n\n\n\n
Correct Answer:<\/strong> The pH of the saturated solution is 0.67<\/strong>.<\/p>\n\n\n\n
Explanation<\/h3>\n\n\n\n
This problem requires determining the pH of a saturated solution of sodium bisulfate (NaHSO\u2084), which involves a weak acid equilibrium calculation.<\/p>\n\n\n\n
1. Calculate the Molar Concentration of the Saturated Solution:<\/strong>
First, the initial concentration of NaHSO\u2084 in the saturated solution is determined by converting the given solubility from grams per liter (g\/L) to moles per liter (mol\/L or M).<\/p>\n\n\n\n
    \n
  • Solubility = 420 g\/L<\/li>\n\n\n\n
  • Molar Mass of NaHSO\u2084 = 104.061\u00a0g\/mol<\/li>\n<\/ul>\n\n\n\n
    Molarity = (420 g\/L) \/ (104.061 g\/mol ) = 4.036 M NaHSO\u2084<\/strong><\/p>\n\n\n\n
    2. Determine the Initial Concentration of the Acidic Ion:<\/strong>
    Sodium bisulfate is an ionic salt that dissociates completely in water into sodium ions (Na\u207a) and bisulfate ions (HSO\u2084\u207b). The sodium ion is a spectator ion and does not affect the pH. The bisulfate ion, however, is a weak acid.<\/p>\n\n\n\n
    NaHSO\u2084(aq) \u2192 Na\u207a(aq) + HSO\u2084\u207b(aq)<\/p>\n\n\n\n
    Due to the 1:1 stoichiometry, the initial concentration of the bisulfate ion, [HSO\u2084\u207b], is equal to the molarity of the solution, which is 4.036 M.<\/p>\n\n\n\n
    3. Set Up the Weak Acid Equilibrium:<\/strong>
    The bisulfate ion (HSO\u2084\u207b) acts as a weak acid, donating a proton to water to establish an equilibrium with sulfate ions (SO\u2084\u00b2\u207b) and hydronium ions (H\u2083O\u207a).<\/p>\n\n\n\n
    The equilibrium reaction is:
    HSO\u2084\u207b(aq) + H\u2082O(l) \u21cc SO\u2084\u00b2\u207b(aq) + H\u2083O\u207a(aq)<\/p>\n\n\n\n
    An ICE (Initial, Change, Equilibrium) table is used to find the equilibrium concentrations:<\/p>\n\n\n\n
    Species<\/td>Initial (M)<\/td>Change (M)<\/td>Equilibrium (M)<\/td><\/tr>
    HSO\u2084\u207b<\/td>4.036<\/td>-x<\/td>4.036 – x<\/td><\/tr>
    SO\u2084\u00b2\u207b<\/td>0<\/td>+x<\/td>x<\/td><\/tr>
    H\u2083O\u207a<\/td>0<\/td>+x<\/td>x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
    4. Solve for the Hydronium Ion Concentration [H\u2083O\u207a]:<\/strong>
    The acid dissociation constant, Ka, is calculated from the given pKa:<\/p>\n\n\n\n
      \n
    • pKa = 1.92<\/li>\n\n\n\n
    • Ka = 10\u207b\u1d56\u1d37\u1d43 = 10\u207b\u00b9\u2079\u00b2 =\u00a00.0120<\/strong><\/li>\n<\/ul>\n\n\n\n
      The equilibrium expression is:
      Ka = [SO\u2084\u00b2\u207b][H\u2083O\u207a] \/ [HSO\u2084\u207b]
      0.0120 = (x)(x) \/ (4.036 – x)<\/p>\n\n\n\n
      Because the initial concentration (4.036 M) is not significantly larger than the Ka value (the ratio C\/Ka is less than 400), the simplifying assumption that ‘x’ is negligible cannot be used. The full quadratic equation must be solved:<\/p>\n\n\n\n
      x\u00b2 = 0.0120 * (4.036 – x)
      x\u00b2 = 0.04843 – 0.0120x
      x\u00b2 + 0.0120x – 0.04843 = 0<\/p>\n\n\n\n
      Using the quadratic formula, x = [-b \u00b1 \u221a(b\u00b2-4ac)] \/ 2a:
      x = [-0.0120 \u00b1 \u221a((0.0120)\u00b2 – 4(1)(-0.04843))] \/ 2
      x = [-0.0120 \u00b1 \u221a(0.000144 + 0.19372)] \/ 2
      x = [-0.0120 \u00b1 \u221a(0.19386)] \/ 2
      x = [-0.0120 \u00b1 0.4403] \/ 2<\/p>\n\n\n\n
      Since concentration cannot be negative, the positive root is taken:
      x = (0.4283) \/ 2 = 0.214 M<\/p>\n\n\n\n
      Thus, the equilibrium concentration of hydronium ions is [H\u2083O\u207a] = 0.214 M<\/strong>.<\/p>\n\n\n\n
      5. Calculate the pH:<\/strong>
      The pH is the negative logarithm of the hydronium ion concentration.<\/p>\n\n\n\n
      pH = -log[H\u2083O\u207a]
      pH = -log(0.214)
      pH \u2248 0.67<\/strong>thumb_upthumb_down<\/p>\n\n\n\n
      \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
      Based on the red ink corrections, solve. The maximum solubility of NaHSO4 is 420 g\/L @ 25\u00b0C. Its molar mass is 104.061 g\/mol. The pKa of HSO4^- is 1.92. What is the pH of a saturated solution of NaHSO4? (6 pts.) Plan: 1.92 The maximum solubility of NaHSO4 is 420 g\/L @ 25\u00b0C. Its molar […]<\/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-234144","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234144","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=234144"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/234144\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=234144"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=234144"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=234144"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}