To calculate the pH of a 0.27 M solution of nitrous acid (HNO2), we need to use the acid dissociation constant (Ka) for HNO2, which is given as 4.0\u00d710\u221244.0 \\times 10^{-4}4.0\u00d710\u22124. Here’s how we can approach the problem:<\/p>\n\n\n\n
Step 1: Write the dissociation equation<\/h3>\n\n\n\n
The dissociation of nitrous acid in water can be written as:HNO2(aq)\u21ccH+(aq)+NO2\u2212(aq)\\text{HNO}_2 (aq) \\rightleftharpoons \\text{H}^+ (aq) + \\text{NO}_2^- (aq)HNO2\u200b(aq)\u21ccH+(aq)+NO2\u2212\u200b(aq)<\/p>\n\n\n\n
The equilibrium expression for the dissociation is:Ka=[H+][NO2\u2212][HNO2]K_a = \\frac{[\\text{H}^+][\\text{NO}_2^-]}{[\\text{HNO}_2]}Ka\u200b=[HNO2\u200b][H+][NO2\u2212\u200b]\u200b<\/p>\n\n\n\n
Given Ka=4.0\u00d710\u22124K_a = 4.0 \\times 10^{-4}Ka\u200b=4.0\u00d710\u22124, we will use the following assumptions:<\/p>\n\n\n\n
- \n
- The concentration of HNO2\\text{HNO}_2HNO2\u200b initially is 0.27 M.<\/li>\n\n\n\n
- The concentration of H+\\text{H}^+H+ and NO2\u2212\\text{NO}_2^-NO2\u2212\u200b will be the same at equilibrium because they are produced in equal amounts.<\/li>\n<\/ul>\n\n\n\n
Step 2: Set up the equilibrium concentrations<\/h3>\n\n\n\n
Let the concentration of H+\\text{H}^+H+ (and NO2\u2212\\text{NO}_2^-NO2\u2212\u200b) at equilibrium be xxx. Initially, the concentration of HNO2\\text{HNO}_2HNO2\u200b is 0.27 M, and it decreases by xxx due to dissociation. So, at equilibrium:<\/p>\n\n\n\n
- \n
- [HNO2]=0.27\u2212x[\\text{HNO}_2] = 0.27 – x[HNO2\u200b]=0.27\u2212x<\/li>\n\n\n\n
- [H+]=x[\\text{H}^+] = x[H+]=x<\/li>\n\n\n\n
- [NO2\u2212]=x[\\text{NO}_2^-] = x[NO2\u2212\u200b]=x<\/li>\n<\/ul>\n\n\n\n
Step 3: Apply the equilibrium expression<\/h3>\n\n\n\n
Substitute the equilibrium concentrations into the expression for KaK_aKa\u200b:4.0\u00d710\u22124=x20.27\u2212x4.0 \\times 10^{-4} = \\frac{x^2}{0.27 – x}4.0\u00d710\u22124=0.27\u2212xx2\u200b<\/p>\n\n\n\n
Assume that xxx is much smaller than 0.27 (since KaK_aKa\u200b is relatively small), so we approximate 0.27\u2212x\u22480.270.27 – x \\approx 0.270.27\u2212x\u22480.27. This simplifies the equation to:4.0\u00d710\u22124=x20.274.0 \\times 10^{-4} = \\frac{x^2}{0.27}4.0\u00d710\u22124=0.27×2\u200b<\/p>\n\n\n\n
Step 4: Solve for xxx<\/h3>\n\n\n\n
Now, solve for xxx:x2=(4.0\u00d710\u22124)(0.27)x^2 = (4.0 \\times 10^{-4})(0.27)x2=(4.0\u00d710\u22124)(0.27)x2=1.08\u00d710\u22124x^2 = 1.08 \\times 10^{-4}x2=1.08\u00d710\u22124x=1.08\u00d710\u22124=0.0104\u2009Mx = \\sqrt{1.08 \\times 10^{-4}} = 0.0104 \\, \\text{M}x=1.08\u00d710\u22124\u200b=0.0104M<\/p>\n\n\n\n
Step 5: Calculate the pH<\/h3>\n\n\n\n
The concentration of H+\\text{H}^+H+ is x=0.0104\u2009Mx = 0.0104 \\, \\text{M}x=0.0104M. The pH is given by:pH=\u2212log\u2061[H+]\\text{pH} = -\\log[\\text{H}^+]pH=\u2212log[H+]pH=\u2212log\u2061(0.0104)=1.98\\text{pH} = -\\log(0.0104) = 1.98pH=\u2212log(0.0104)=1.98<\/p>\n\n\n\n
Conclusion<\/h3>\n\n\n\n
The pH of the 0.27 M HNO2 solution is approximately 1.98<\/strong>.<\/p>\n\n\n\n
This calculation assumes that the dissociation is relatively weak, which is true for many weak acids like HNO2. The Ka value indicates that only a small fraction of the acid dissociates, so the pH is relatively low but not as acidic as a strong acid like HCl, for example.<\/p>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"For the nitrous acid, HNO2, Ka= 4.0 x 10^-4. Calculate the ph of 0.27 M HNO2 The Correct Answer and Explanation is: To calculate the pH of a 0.27 M solution of nitrous acid (HNO2), we need to use the acid dissociation constant (Ka) for HNO2, which is given as 4.0\u00d710\u221244.0 \\times 10^{-4}4.0\u00d710\u22124. Here’s how […]<\/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-244712","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244712","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=244712"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/244712\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=244712"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=244712"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=244712"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}