Calculate the pH of a 0.33 M nitrous acid (HNO2, Ka = 4.5 x10-4) solution.
0.48
3.82
1.92
None is correct<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the pH of a 0.33 M nitrous acid (HNO\u2082) solution, we use the following steps:<\/p>\n\n\n\n Nitrous acid (HNO\u2082) is a weak acid and partially dissociates in water as follows:<\/p>\n\n\n\n [ Given that the acid dissociation constant (( K_a )) for nitrous acid is:<\/p>\n\n\n\n [ Let\u2019s define the initial concentration and the changes due to dissociation:<\/p>\n\n\n\n [ Substituting the equilibrium values:<\/p>\n\n\n\n [ Since ( K_a ) is small, we assume ( x ) is small compared to 0.33, so we approximate:<\/p>\n\n\n\n [ Thus, the equation simplifies to:<\/p>\n\n\n\n [ [ [ [ [ Since ( x = [H^+] ), we find:<\/p>\n\n\n\n [ [ The closest option is none of the given answers<\/strong> because 3.92 is not listed.<\/p>\n\n\n\n The pH of a 0.33 M nitrous acid solution was determined using the acid dissociation constant (( K_a )) and the ICE table method. Since nitrous acid is a weak acid, it does not completely dissociate in solution. We set up an equilibrium expression using the ( K_a ) value and approximated the dissociation to simplify calculations.<\/p>\n\n\n\n By solving for ( [H^+] ), we found it to be approximately 0.0122 M. Taking the negative logarithm of the hydrogen ion concentration gave us a pH of 3.92<\/strong>, which was not listed among the answer choices.<\/p>\n\n\n\n This calculation highlights an important principle in acid-base chemistry: weak acids partially ionize, and their pH must be determined using equilibrium expressions rather than assuming full dissociation. Approximations are often used in these calculations when ( K_a ) is small compared to the initial concentration, as in this case.<\/p>\n\n\n\n This result confirms that none of the provided answer choices are correct. If an answer choice of 3.92<\/strong> had been available, it would have been the correct selection.<\/p>\n","protected":false},"excerpt":{"rendered":" Calculate the pH of a 0.33 M nitrous acid (HNO2, Ka = 4.5 x10-4) solution.0.483.821.92None is correct The Correct Answer and Explanation is : To calculate the pH of a 0.33 M nitrous acid (HNO\u2082) solution, we use the following steps: Step 1: Write the Dissociation Equation Nitrous acid (HNO\u2082) is a weak acid 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-190874","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190874","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=190874"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190874\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190874"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190874"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190874"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Write the Dissociation Equation<\/h3>\n\n\n\n
HNO_2 \\rightleftharpoons H^+ + NO_2^-
]<\/p>\n\n\n\n
K_a = 4.5 \\times 10^{-4}
]<\/p>\n\n\n\nStep 2: Set Up the ICE Table<\/h3>\n\n\n\n
Species<\/th> Initial (M)<\/th> Change (M)<\/th> Equilibrium (M)<\/th><\/tr><\/thead> HNO\u2082<\/td> 0.33<\/td> -x<\/td> 0.33 – x<\/td><\/tr> H\u207a<\/td> 0<\/td> +x<\/td> x<\/td><\/tr> NO\u2082\u207b<\/td> 0<\/td> +x<\/td> x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Step 3: Apply the ( K_a ) Expression<\/h3>\n\n\n\n
K_a = \\frac{[H^+][NO_2^-]}{[HNO_2]}
]<\/p>\n\n\n\n
4.5 \\times 10^{-4} = \\frac{x^2}{0.33 – x}
]<\/p>\n\n\n\n
0.33 – x \\approx 0.33
]<\/p>\n\n\n\n
4.5 \\times 10^{-4} = \\frac{x^2}{0.33}
]<\/p>\n\n\n\nStep 4: Solve for ( x )<\/h3>\n\n\n\n
x^2 = (4.5 \\times 10^{-4}) \\times (0.33)
]<\/p>\n\n\n\n
x^2 = 1.485 \\times 10^{-4}
]<\/p>\n\n\n\n
x = \\sqrt{1.485 \\times 10^{-4}}
]<\/p>\n\n\n\n
x \\approx 0.0122
]<\/p>\n\n\n\n
pH = -\\log(0.0122)
]<\/p>\n\n\n\n
pH \\approx 3.92
]<\/p>\n\n\n\nStep 5: Compare with Given Answers<\/h3>\n\n\n\n
Explanation (300 words)<\/h3>\n\n\n\n