Determine the pH of a .22 M NaF solution. The Ka of HF is 3.5 x 10^-5<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the pH of a 0.22 M NaF solution, we need to consider the hydrolysis of fluoride ions (F\u207b) in water. Sodium fluoride (NaF) is a salt that dissociates completely in water into Na\u207a and F\u207b. Since Na\u207a is a spectator ion, the fluoride ion undergoes hydrolysis:<\/p>\n\n\n\n [ The equilibrium constant for this reaction is the base dissociation constant ( K_b ), which is related to the acid dissociation constant ( K_a ) of HF by:<\/p>\n\n\n\n [ where:<\/p>\n\n\n\n [ [ Let ( x ) be the concentration of OH\u207b produced:<\/p>\n\n\n\n [ [ Solving for ( x ):<\/p>\n\n\n\n [ [ [ [ [ [ [ The pH of a 0.22 M NaF solution is 8.90<\/strong>.<\/p>\n\n\n\n Sodium fluoride is a basic salt because fluoride is the conjugate base of HF. When NaF dissolves, F\u207b hydrolyzes in water to produce OH\u207b, making the solution basic. We determined the pH by calculating the ( K_b ) using the relationship with ( K_a ), solving for hydroxide concentration, and converting to pH. The result confirms that NaF solutions are mildly basic.<\/p>\n","protected":false},"excerpt":{"rendered":" Determine the pH of a .22 M NaF solution. The Ka of HF is 3.5 x 10^-5 The Correct Answer and Explanation is : To determine the pH of a 0.22 M NaF solution, we need to consider the hydrolysis of fluoride ions (F\u207b) in water. Sodium fluoride (NaF) is a salt that dissociates completely […]<\/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-187312","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187312","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=187312"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187312\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187312"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187312"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187312"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
F^- + H_2O \\rightleftharpoons HF + OH^-
]<\/p>\n\n\n\nStep 1: Determine the ( K_b ) of ( F^- )<\/h3>\n\n\n\n
K_b = \\frac{K_w}{K_a}
]<\/p>\n\n\n\n\n
K_b = \\frac{1.0 \\times 10^{-14}}{3.5 \\times 10^{-5}}
]<\/p>\n\n\n\n
K_b = 2.86 \\times 10^{-10}
]<\/p>\n\n\n\nStep 2: Set Up the ICE Table for Hydrolysis<\/h3>\n\n\n\n
K_b = \\frac{x^2}{0.22 – x} \\approx \\frac{x^2}{0.22}
]<\/p>\n\n\n\n
2.86 \\times 10^{-10} = \\frac{x^2}{0.22}
]<\/p>\n\n\n\n
x = \\sqrt{(2.86 \\times 10^{-10}) \\times 0.22}
]<\/p>\n\n\n\n
x = \\sqrt{6.292 \\times 10^{-11}}
]<\/p>\n\n\n\n
x = 7.94 \\times 10^{-6} \\text{ M} \\quad (\\text{[OH\u207b]})
]<\/p>\n\n\n\nStep 3: Determine pOH and pH<\/h3>\n\n\n\n
\\text{pOH} = -\\log(7.94 \\times 10^{-6})
]<\/p>\n\n\n\n
\\text{pOH} \\approx 5.10
]<\/p>\n\n\n\n
\\text{pH} = 14 – 5.10
]<\/p>\n\n\n\n
\\text{pH} \\approx 8.90
]<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n