{"id":190871,"date":"2025-02-13T10:05:03","date_gmt":"2025-02-13T10:05:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190871"},"modified":"2025-02-13T10:05:07","modified_gmt":"2025-02-13T10:05:07","slug":"for-nitrous-acid-hno2-ka-4-0-x-10-4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/13\/for-nitrous-acid-hno2-ka-4-0-x-10-4\/","title":{"rendered":"For nitrous acid, HNO2, Ka = 4.0 x 10^-4"},"content":{"rendered":"\n

For nitrous acid, HNO2, Ka = 4.0 x 10^-4. Calculate the pH of 0.54 M HNO2. a. 1.83
b. 0.27
C. 3.67
d. 12.17
e. none of these<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

To calculate the pH of a 0.54 M solution of nitrous acid (HNO\u2082), we follow these steps:<\/p>\n\n\n\n

Step 1: Set Up the ICE Table<\/strong><\/h3>\n\n\n\n

Nitrous acid dissociates in water as follows:<\/p>\n\n\n\n

[
HNO_2 \\rightleftharpoons H^+ + NO_2^-
]<\/p>\n\n\n\n

Let\u2019s define the initial concentration of HNO\u2082 as 0.54 M<\/strong>, and the initial concentrations of H\u207a and NO\u2082\u207b as 0 M<\/strong>.<\/p>\n\n\n\n

Species<\/th>Initial (M)<\/th>Change (M)<\/th>Equilibrium (M)<\/th><\/tr><\/thead>
HNO\u2082<\/td>0.54<\/td>-x<\/td>0.54 – 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 2: Write the Ka Expression<\/strong><\/h3>\n\n\n\n

[
Ka = \\frac{[H^+][NO_2^-]}{[HNO_2]}
]<\/p>\n\n\n\n

Substituting the given values:<\/p>\n\n\n\n

[
4.0 \\times 10^{-4} = \\frac{x \\cdot x}{0.54 – x}
]<\/p>\n\n\n\n

Since ( x ) is small compared to 0.54, we approximate 0.54 – x \u2248 0.54<\/strong>:<\/p>\n\n\n\n

[
4.0 \\times 10^{-4} = \\frac{x^2}{0.54}
]<\/p>\n\n\n\n

Step 3: Solve for x<\/strong><\/h3>\n\n\n\n

[
x^2 = (4.0 \\times 10^{-4}) (0.54)
]<\/p>\n\n\n\n

[
x^2 = 2.16 \\times 10^{-4}
]<\/p>\n\n\n\n

[
x = \\sqrt{2.16 \\times 10^{-4}}
]<\/p>\n\n\n\n

[
x = 0.0147 \\text{ M}
]<\/p>\n\n\n\n

Since x represents [H\u207a]<\/strong>, we find the pH:<\/p>\n\n\n\n

[
pH = -\\log (0.0147)
]<\/p>\n\n\n\n

[
pH = 3.67
]<\/p>\n\n\n\n

Final Answer:<\/strong><\/h3>\n\n\n\n

[
\\mathbf{3.67} \\quad (C)
]<\/p>\n\n\n\n

\n\n\n\n

Explanation (300 Words)<\/strong><\/h3>\n\n\n\n

Nitrous acid (HNO\u2082) is a weak acid, meaning it only partially dissociates in water. The acid dissociation constant (Ka) is 4.0 \u00d7 10\u207b\u2074<\/strong>, which helps determine the extent of ionization. To find the pH of a 0.54 M<\/strong> solution of HNO\u2082, we use an ICE table<\/strong> (Initial, Change, Equilibrium) to track the concentration changes.<\/p>\n\n\n\n

Initially, HNO\u2082 is present at 0.54 M<\/strong>, while the concentrations of H\u207a and NO\u2082\u207b<\/strong> are 0 M<\/strong>. As the acid dissociates, some molecules break into H\u207a (protons) and NO\u2082\u207b (conjugate base)<\/strong>. We define the concentration of H\u207a formed as x<\/strong> and set up the equilibrium equation:<\/p>\n\n\n\n

[
Ka = \\frac{x^2}{0.54 – x}
]<\/p>\n\n\n\n

Since Ka is small, x is small compared to 0.54<\/strong>, allowing the approximation 0.54 – x \u2248 0.54<\/strong> for easier calculations. Solving for x<\/strong>, we get 0.0147 M<\/strong>, which represents the concentration of hydrogen ions in solution.<\/p>\n\n\n\n

Finally, we calculate pH = -log[H\u207a]<\/strong>, giving pH = 3.67<\/strong>. The correct answer is option C<\/strong>. This makes sense because nitrous acid is a weak acid, meaning its pH should be greater than 1 but less than 7<\/strong>.<\/p>\n\n\n\n

This approach is widely used in chemistry to determine the pH of weak acids, especially in buffer and equilibrium calculations. Approximations simplify the process, but for very diluted solutions, a quadratic equation may be necessary.<\/p>\n\n\n\n

Now, let’s generate an image illustrating this calculation.<\/p>\n\n\n\n

Here is the educational chemistry diagram illustrating the pH calculation of a 0.54 M solution of nitrous acid (HNO\u2082). The image includes an ICE table, the Ka expression, and a step-by-step solution leading to the final pH value of 3.67. Let me know if you need any modifications or further explanations!<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

For nitrous acid, HNO2, Ka = 4.0 x 10^-4. Calculate the pH of 0.54 M HNO2. a. 1.83b. 0.27C. 3.67d. 12.17e. none of these The Correct Answer and Explanation is : To calculate the pH of a 0.54 M solution of nitrous acid (HNO\u2082), we follow these steps: Step 1: Set Up the ICE Table […]<\/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-190871","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190871","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=190871"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190871\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190871"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190871"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190871"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}