Working with Aqueous Solutions of HF and NaF<\/p>\n\n\n\n
a. What is the pH of a 0.35 M hydrofluoric acid (HF) solution? Ka 6.8x 104 [1.81]<\/p>\n\n\n\n
b. What is the pH of a solution containing 0.35 M HF and 0.28 M NaF? [3.07]<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s go step by step for both calculations.<\/p>\n\n\n\n Hydrofluoric acid (HF) is a weak acid that partially ionizes in water:<\/p>\n\n\n\n [ Given:<\/p>\n\n\n\n Using the ICE (Initial, Change, Equilibrium) table and the expression for ( K_a ):<\/p>\n\n\n\n [ Let ( x ) be the concentration of ( H^+ ) that dissociates. At equilibrium:<\/p>\n\n\n\n [ Since ( K_a ) is small, we approximate ( 0.35 – x \\approx 0.35 ), so:<\/p>\n\n\n\n [ Solving for ( x ):<\/p>\n\n\n\n [ [ [ Thus,<\/p>\n\n\n\n [ This is a buffer solution because it contains both a weak acid (HF) and its conjugate base (( F^- )) from NaF. We use the Henderson-Hasselbalch equation<\/strong>:<\/p>\n\n\n\n [ First, calculate ( pK_a ):<\/p>\n\n\n\n [ Now, plug in the concentrations:<\/p>\n\n\n\n [ [ [ Hydrofluoric acid (HF) is a weak acid because it does not completely dissociate in water. The dissociation is controlled by its acid dissociation constant (( K_a )), which is ( 6.8 \\times 10^{-4} ). To determine the pH of a pure HF solution, we use an ICE table to set up an equilibrium equation based on ( K_a ). Since ( K_a ) is relatively small, we approximate the equilibrium expression, solving for ( x ), the ( H^+ ) concentration, to find the pH of 1.81.<\/p>\n\n\n\n When sodium fluoride (NaF) is added, it dissociates completely into ( Na^+ ) and ( F^- ). The presence of fluoride ions suppresses the dissociation of HF due to Le Chatelier\u2019s principle. This creates a buffer system, where HF (weak acid) and ( F^- ) (conjugate base) resist changes in pH.<\/p>\n\n\n\n To calculate the pH of the buffer, we apply the Henderson-Hasselbalch equation<\/strong>, which is derived from the acid dissociation expression. The equation relates pH to the ( pK_a ) and the ratio of the conjugate base (( F^- )) to the weak acid (HF). Substituting given concentrations, we find the pH to be 3.07.<\/p>\n\n\n\n Buffers are crucial in biological and chemical systems as they stabilize pH. This HF\/NaF buffer system, for example, is used in fluoride treatments and industrial applications.<\/p>\n","protected":false},"excerpt":{"rendered":" Working with Aqueous Solutions of HF and NaF a. What is the pH of a 0.35 M hydrofluoric acid (HF) solution? Ka 6.8x 104 [1.81] b. What is the pH of a solution containing 0.35 M HF and 0.28 M NaF? [3.07] The Correct Answer and Explanation is : Let’s go step by step for […]<\/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-187314","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187314","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=187314"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187314\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187314"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187314"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187314"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}(a) Finding the pH of a 0.35 M HF Solution<\/strong><\/h3>\n\n\n\n
HF \\rightleftharpoons H^+ + F^-
]<\/p>\n\n\n\n\n
K_a = \\frac{[H^+][F^-]}{[HF]}
]<\/p>\n\n\n\n
K_a = \\frac{x^2}{0.35 – x}
]<\/p>\n\n\n\n
6.8 \\times 10^{-4} = \\frac{x^2}{0.35}
]<\/p>\n\n\n\n
x^2 = (6.8 \\times 10^{-4}) \\times (0.35)
]<\/p>\n\n\n\n
x^2 = 2.38 \\times 10^{-4}
]<\/p>\n\n\n\n
x = \\sqrt{2.38 \\times 10^{-4}} = 0.0154
]<\/p>\n\n\n\n
\\text{pH} = -\\log(0.0154) = 1.81
]<\/p>\n\n\n\n(b) Finding the pH of a Buffer Solution (0.35 M HF and 0.28 M NaF)<\/strong><\/h3>\n\n\n\n
\\text{pH} = pK_a + \\log \\frac{[A^-]}{[HA]}
]<\/p>\n\n\n\n
pK_a = -\\log(6.8 \\times 10^{-4}) = 3.17
]<\/p>\n\n\n\n
\\text{pH} = 3.17 + \\log \\frac{0.28}{0.35}
]<\/p>\n\n\n\n
\\text{pH} = 3.17 + \\log(0.8)
]<\/p>\n\n\n\n
\\text{pH} = 3.17 + (-0.10) = 3.07
]<\/p>\n\n\n\nExplanation (300 Words)<\/strong><\/h3>\n\n\n\n