To calculate the pH of a 0.1 M NaF solution, we need to consider the dissociation of NaF in water and how the fluoride ion (F\u207b) behaves in solution.<\/p>\n\n\n\n
- \n
- Dissociation of NaF<\/strong>: Sodium fluoride (NaF) dissociates completely in water:<\/li>\n<\/ol>\n\n\n\n
[
\\text{NaF} \\rightarrow \\text{Na}^+ + \\text{F}^-
]<\/p>\n\n\n\nSo, the concentration of fluoride ions (F\u207b) in the solution is 0.1 M.<\/p>\n\n\n\n
- \n
- Behavior of fluoride ions in water<\/strong>: Fluoride ions act as a base and can react with water to produce hydroxide ions (OH\u207b) according to the following reaction:<\/li>\n<\/ol>\n\n\n\n
[
\\text{F}^- + \\text{H}_2\\text{O} \\rightleftharpoons \\text{HF} + \\text{OH}^-
]<\/p>\n\n\n\n- \n
- Equilibrium expression<\/strong>: The equilibrium constant for this reaction, ( K_b ), can be calculated using the relationship between ( K_b ) and ( K_a ) (the ionization constant for HF):<\/li>\n<\/ol>\n\n\n\n
[
K_w = K_a \\times K_b
]<\/p>\n\n\n\nwhere ( K_w ) is the ionization constant for water (( K_w = 1 \\times 10^{-14} )) and ( K_a ) is given as ( 7 \\times 10^{-4} ) for HF.<\/p>\n\n\n\n
We can solve for ( K_b ):<\/p>\n\n\n\n
[
K_b = \\frac{K_w}{K_a} = \\frac{1 \\times 10^{-14}}{7 \\times 10^{-4}} = 1.43 \\times 10^{-11}
]<\/p>\n\n\n\n- \n
- Calculating OH\u207b concentration<\/strong>: The equilibrium expression for the dissociation of F\u207b in water is:<\/li>\n<\/ol>\n\n\n\n
[
K_b = \\frac{[\\text{HF}][\\text{OH}^-]}{[\\text{F}^-]}
]<\/p>\n\n\n\nLet ( x ) represent the concentration of OH\u207b produced at equilibrium. The equilibrium concentrations would be:<\/p>\n\n\n\n
- \n
- ([F^-] = 0.1 – x)<\/li>\n\n\n\n
- ([HF] = x)<\/li>\n\n\n\n
- ([OH^-] = x)<\/li>\n<\/ul>\n\n\n\n
Substituting into the equilibrium expression:<\/p>\n\n\n\n
[
1.43 \\times 10^{-11} = \\frac{x^2}{0.1}
]<\/p>\n\n\n\nSolving for ( x ) (concentration of OH\u207b):<\/p>\n\n\n\n
[
x^2 = (1.43 \\times 10^{-11}) \\times 0.1 = 1.43 \\times 10^{-12}
]<\/p>\n\n\n\n[
x = \\sqrt{1.43 \\times 10^{-12}} = 1.19 \\times 10^{-6} \\, \\text{M}
]<\/p>\n\n\n\nThis is the concentration of OH\u207b.<\/p>\n\n\n\n
- \n
- Calculating pOH<\/strong>: The pOH is given by:<\/li>\n<\/ol>\n\n\n\n
[
\\text{pOH} = -\\log[\\text{OH}^-] = -\\log(1.19 \\times 10^{-6}) = 5.92
]<\/p>\n\n\n\n- \n
- Calculating pH<\/strong>: The pH is related to the pOH by the equation:<\/li>\n<\/ol>\n\n\n\n
[
\\text{pH} + \\text{pOH} = 14
]<\/p>\n\n\n\nThus,<\/p>\n\n\n\n
[
\\text{pH} = 14 – 5.92 = 8.08
]<\/p>\n\n\n\nTherefore, the pH of the 0.1 M NaF solution is approximately 8.1<\/strong>.<\/p>\n\n\n\n
Answer: 8.1<\/strong>.<\/p>\n\n\n\n
I will now generate a visual explanation of the calculation.<\/p>\n\n\n\n
Here is a detailed diagram illustrating the process of calculating the pH of a 0.1 M NaF solution. It covers each step from dissociation to the final pH calculation. Let me know if you need further clarification!<\/p>\n\n\n\n
![\"\"](\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-1078.png\")
<\/figure>\n","protected":false},"excerpt":{"rendered":"What is the pH of a 0.1 M NaF solution? Ionization constant for HF, Ka= 7 x 102.15.98.19.1 The Correct Answer and Explanation is : To calculate the pH of a 0.1 M NaF solution, we need to consider the dissociation of NaF in water and how the fluoride ion (F\u207b) behaves in solution. [\\text{NaF} […]<\/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-193604","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193604","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=193604"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193604\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193604"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193604"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193604"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Calculating pH<\/strong>: The pH is related to the pOH by the equation:<\/li>\n<\/ol>\n\n\n\n
- Calculating pOH<\/strong>: The pOH is given by:<\/li>\n<\/ol>\n\n\n\n
- Calculating OH\u207b concentration<\/strong>: The equilibrium expression for the dissociation of F\u207b in water is:<\/li>\n<\/ol>\n\n\n\n
- Equilibrium expression<\/strong>: The equilibrium constant for this reaction, ( K_b ), can be calculated using the relationship between ( K_b ) and ( K_a ) (the ionization constant for HF):<\/li>\n<\/ol>\n\n\n\n
- Behavior of fluoride ions in water<\/strong>: Fluoride ions act as a base and can react with water to produce hydroxide ions (OH\u207b) according to the following reaction:<\/li>\n<\/ol>\n\n\n\n