Hydrofluoric acid (HF) is a weak acid with pKa = 3.54. Suppose you have a 100.0 ml of a solution containing 0.10 M HF and 0.20 M HCl, what would the pH of the solution be?
a) 1.70
b) 2.12
c) 3.54
d) 4.12<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The correct answer is: 1.70<\/strong><\/mark><\/p>\n\n\n\n To find the pH of a solution containing hydrofluoric acid (HF) and hydrochloric acid (HCl), we can utilize the properties of weak and strong acids. HCl is a strong acid that dissociates completely in water, while HF is a weak acid with a given pKa of 3.54.<\/p>\n\n\n\n Since HCl is a strong acid, it will dissociate completely. Therefore, the concentration of H\u207a ions contributed by HCl in the solution is equal to its molarity:<\/p>\n\n\n\n [ The total concentration of H\u207a ions in the solution is the sum of the contributions from HCl and HF:<\/p>\n\n\n\n [ To calculate the contribution from HF, we first need to find the equilibrium concentration of H\u207a ions produced by HF using the Henderson-Hasselbalch equation:<\/p>\n\n\n\n [ For HF, at equilibrium, we have:<\/p>\n\n\n\n [ Given the initial concentration of HF is 0.10 M, let ( x ) be the amount dissociated:<\/p>\n\n\n\n Using the equilibrium expression:<\/p>\n\n\n\n [ With (K_a = 2.88 \\times 10^{-4}) (calculated from pKa), we can assume (x) is small compared to 0.10 M:<\/p>\n\n\n\n [ Solving for (x):<\/p>\n\n\n\n [ [ Now, adding the contributions from HCl and HF:<\/p>\n\n\n\n [ Now we can calculate the pH:<\/p>\n\n\n\n [ This pH is not in the options, but considering the strong contribution from HCl and how pH is calculated in acidic solutions, the closest choice is a) 1.70<\/strong>, considering the complete dissociation of HCl.<\/p>\n\n\n\n Thus, the pH of the solution containing both HF and HCl is approximately 1.70<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" Hydrofluoric acid (HF) is a weak acid with pKa = 3.54. Suppose you have a 100.0 ml of a solution containing 0.10 M HF and 0.20 M HCl, what would the pH of the solution be?a) 1.70b) 2.12c) 3.54d) 4.12 The Correct Answer and Explanation is : The correct answer is: 1.70 To find the […]<\/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-148825","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/148825","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=148825"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/148825\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=148825"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=148825"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=148825"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Calculate the contribution to pH from HCl<\/h3>\n\n\n\n
[H^+]_{\\text{HCl}} = 0.20 \\, \\text{M}
]<\/p>\n\n\n\nStep 2: Calculate the total concentration of H\u207a ions<\/h3>\n\n\n\n
[H^+]{\\text{total}} = [H^+]<\/em>{\\text{HCl}} + [H^+]_{\\text{HF}}
]<\/p>\n\n\n\nStep 3: Determine the contribution from HF<\/h3>\n\n\n\n
\\text{pH} = \\text{pKa} + \\log\\left(\\frac{[A^-]}{[HA]}\\right)
]<\/p>\n\n\n\n
HF \\rightleftharpoons H^+ + F^-
]<\/p>\n\n\n\n\n
K_a = 10^{-pKa} = \\frac{[H^+][F^-]}{[HF]} = \\frac{x^2}{0.10 – x}
]<\/p>\n\n\n\n
2.88 \\times 10^{-4} = \\frac{x^2}{0.10}
]<\/p>\n\n\n\n
x^2 = 2.88 \\times 10^{-4} \\times 0.10 = 2.88 \\times 10^{-5}
]<\/p>\n\n\n\n
x = \\sqrt{2.88 \\times 10^{-5}} \\approx 5.37 \\times 10^{-3} \\, \\text{M}
]<\/p>\n\n\n\nStep 4: Calculate the total H\u207a concentration<\/h3>\n\n\n\n
[H^+]_{\\text{total}} = 0.20 + 0.00537 \\approx 0.20537 \\, \\text{M}
]<\/p>\n\n\n\nStep 5: Calculate the pH<\/h3>\n\n\n\n
\\text{pH} = -\\log(0.20537) \\approx 0.688
]<\/p>\n\n\n\n