{"id":262808,"date":"2025-07-20T19:09:56","date_gmt":"2025-07-20T19:09:56","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=262808"},"modified":"2025-07-20T19:09:58","modified_gmt":"2025-07-20T19:09:58","slug":"an-aqueous-solution-of-naf-is-prepared-by-dissolving-0-350-mol-of-naf-in-sufficient-water-to-yield-1-0-l-of-solution","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/20\/an-aqueous-solution-of-naf-is-prepared-by-dissolving-0-350-mol-of-naf-in-sufficient-water-to-yield-1-0-l-of-solution\/","title":{"rendered":"An aqueous solution of NaF is prepared by dissolving 0.350 mol of NaF in sufficient water to yield 1.0 L of solution."},"content":{"rendered":"\n

The pH of the solution was 8.93 at 25.0 \u00c2\u00b0C. Kw = 1.00 x 10^-14 M^2. Numbers with exponents like Kw must be input as -1.00e-14. {As usual, pay attention to significant figures.} Determine the following: The concentration of HF at equilibrium, the Kb of F^-, and the Ka of HF.<\/p>\n\n\n\n

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

To solve this problem, let’s break it down step by step:<\/p>\n\n\n\n

Step 1: Understanding the Solution<\/h3>\n\n\n\n

We have a 0.350 mol\/L solution of NaF. When NaF dissociates in water, it forms Na\u207a and F\u207b ions:
NaF\u2192Na++F\u2212\\text{NaF} \\rightarrow \\text{Na}^+ + \\text{F}^-NaF\u2192Na++F\u2212<\/p>\n\n\n\n

The pH of the solution is given as 8.93. To find the pOH, we use the following relationship:pH+pOH=14\\text{pH} + \\text{pOH} = 14pH+pOH=14pOH=14\u22128.93=5.07\\text{pOH} = 14 – 8.93 = 5.07pOH=14\u22128.93=5.07<\/p>\n\n\n\n

The concentration of hydroxide ions (OH\u207b) can be calculated using:[OH\u2212]=10\u2212pOH=10\u22125.07=8.51\u00d710\u22126\u2009M[\\text{OH}^-] = 10^{-\\text{pOH}} = 10^{-5.07} = 8.51 \\times 10^{-6} \\, \\text{M}[OH\u2212]=10\u2212pOH=10\u22125.07=8.51\u00d710\u22126M<\/p>\n\n\n\n

Step 2: F\u207b Hydrolysis<\/h3>\n\n\n\n

The fluoride ion (F\u207b) is a weak base and will react with water to form HF and OH\u207b ions:F\u2212+H2O\u21ccHF+OH\u2212\\text{F}^- + \\text{H}_2\\text{O} \\rightleftharpoons \\text{HF} + \\text{OH}^-F\u2212+H2\u200bO\u21ccHF+OH\u2212<\/p>\n\n\n\n

The concentration of OH\u207b ions at equilibrium is 8.51 \u00d7 10\u207b\u2076 M. The reaction begins with 0.350 M of F\u207b (since it comes directly from NaF dissociation).<\/p>\n\n\n\n

Let’s use the equilibrium expression for the base dissociation constant (Kb) of F\u207b:Kb=[HF][OH\u2212][F\u2212]\\text{Kb} = \\frac{[\\text{HF}][\\text{OH}^-]}{[\\text{F}^-]}Kb=[F\u2212][HF][OH\u2212]\u200b<\/p>\n\n\n\n

Step 3: Set Up the ICE Table<\/h3>\n\n\n\n

Let\u2019s assume the change in the concentration of F\u207b is xxx, and the concentrations of HF and OH\u207b at equilibrium are both xxx. Initially, the concentration of F\u207b is 0.350 M.[F\u2212][HF][OH\u2212]Initial (M)0.35000Change (M)\u2212xxxEquilibrium (M)0.350\u2212xx8.51\u00d710\u22126\\begin{array}{|c|c|c|c|} \\hline & [\\text{F}^-] & [\\text{HF}] & [\\text{OH}^-] \\\\ \\hline \\text{Initial (M)} & 0.350 & 0 & 0 \\\\ \\text{Change (M)} & -x & x & x \\\\ \\text{Equilibrium (M)} & 0.350 – x & x & 8.51 \\times 10^{-6} \\\\ \\hline \\end{array}Initial (M)Change (M)Equilibrium (M)\u200b[F\u2212]0.350\u2212x0.350\u2212x\u200b[HF]0xx\u200b[OH\u2212]0x8.51\u00d710\u22126\u200b\u200b<\/p>\n\n\n\n

Since the concentration of OH\u207b at equilibrium is 8.51 \u00d7 10\u207b\u2076 M, we can say that:x=8.51\u00d710\u22126\u2009Mx = 8.51 \\times 10^{-6} \\, \\text{M}x=8.51\u00d710\u22126M<\/p>\n\n\n\n

Step 4: Solving for Kb of F\u207b<\/h3>\n\n\n\n

Now we can substitute these values into the equilibrium expression for Kb:Kb=(8.51\u00d710\u22126)(8.51\u00d710\u22126)0.350\u22128.51\u00d710\u22126\\text{Kb} = \\frac{(8.51 \\times 10^{-6})(8.51 \\times 10^{-6})}{0.350 – 8.51 \\times 10^{-6}}Kb=0.350\u22128.51\u00d710\u22126(8.51\u00d710\u22126)(8.51\u00d710\u22126)\u200b<\/p>\n\n\n\n

Since 8.51\u00d710\u221268.51 \\times 10^{-6}8.51\u00d710\u22126 is much smaller than 0.350, we can approximate:Kb=(8.51\u00d710\u22126)20.350\\text{Kb} = \\frac{(8.51 \\times 10^{-6})^2}{0.350}Kb=0.350(8.51\u00d710\u22126)2\u200bKb=7.24\u00d710\u2212110.350=2.07\u00d710\u221210\u2009M\\text{Kb} = \\frac{7.24 \\times 10^{-11}}{0.350} = 2.07 \\times 10^{-10} \\, \\text{M}Kb=0.3507.24\u00d710\u221211\u200b=2.07\u00d710\u221210M<\/p>\n\n\n\n

Step 5: Finding Ka of HF<\/h3>\n\n\n\n

Next, we can find the Ka of HF using the relationship between Ka and Kb:Ka\u00d7Kb=Kw=1.00\u00d710\u221214\u2009M2\\text{Ka} \\times \\text{Kb} = \\text{Kw} = 1.00 \\times 10^{-14} \\, \\text{M}^2Ka\u00d7Kb=Kw=1.00\u00d710\u221214M2Ka=KwKb=1.00\u00d710\u2212142.07\u00d710\u221210=4.83\u00d710\u22125\u2009M\\text{Ka} = \\frac{\\text{Kw}}{\\text{Kb}} = \\frac{1.00 \\times 10^{-14}}{2.07 \\times 10^{-10}} = 4.83 \\times 10^{-5} \\, \\text{M}Ka=KbKw\u200b=2.07\u00d710\u2212101.00\u00d710\u221214\u200b=4.83\u00d710\u22125M<\/p>\n\n\n\n

Step 6: Concentration of HF at Equilibrium<\/h3>\n\n\n\n

Finally, the concentration of HF at equilibrium is equal to xxx, which is:[HF]=8.51\u00d710\u22126\u2009M[\\text{HF}] = 8.51 \\times 10^{-6} \\, \\text{M}[HF]=8.51\u00d710\u22126M<\/p>\n\n\n\n

Final Answers:<\/h3>\n\n\n\n
    \n
  • The concentration of HF at equilibrium = 8.51\u00d710\u22126\u2009M8.51 \\times 10^{-6} \\, \\text{M}8.51\u00d710\u22126M<\/li>\n\n\n\n
  • The Kb of F\u207b = 2.07\u00d710\u221210\u2009M2.07 \\times 10^{-10} \\, \\text{M}2.07\u00d710\u221210M<\/li>\n\n\n\n
  • The Ka of HF = 4.83\u00d710\u22125\u2009M4.83 \\times 10^{-5} \\, \\text{M}4.83\u00d710\u22125M<\/li>\n<\/ul>\n\n\n\n

    These are the results based on the calculations.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    The pH of the solution was 8.93 at 25.0 \u00c2\u00b0C. Kw = 1.00 x 10^-14 M^2. Numbers with exponents like Kw must be input as -1.00e-14. {As usual, pay attention to significant figures.} Determine the following: The concentration of HF at equilibrium, the Kb of F^-, and the Ka of HF. The Correct Answer and […]<\/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-262808","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/262808","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=262808"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/262808\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=262808"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=262808"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=262808"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}