In the laboratory a student measures the percent ionization of a 0.452 M solution of hypochlorous acid to be 0.0287%. Calculate value of ka from this experimental data.<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n Let’s solve the problem step-by-step:<\/p>\n\n\n\n HOCl\u21ccH++OCl\u2212\\mathrm{HOCl} \\rightleftharpoons \\mathrm{H^+} + \\mathrm{OCl^-}<\/p>\n\n\n\n Percent ionization is given by: %ionization=[H+][HA]initial\u00d7100\\% \\text{ionization} = \\frac{[\\mathrm{H^+}]}{[HA]_{initial}} \\times 100<\/p>\n\n\n\n Given the percent ionization is 0.0287%, convert it to decimal: 0.0287%=0.0002870.0287\\% = 0.000287<\/p>\n\n\n\n Concentration of ionized H+\\mathrm{H^+} or OCl\u2212\\mathrm{OCl^-} is: [H+]=[OCl\u2212]=0.000287\u00d70.452=0.0001297\u2009M[\\mathrm{H^+}] = [\\mathrm{OCl^-}] = 0.000287 \\times 0.452 = 0.0001297\\, M<\/p>\n\n\n\n Ka=[H+][OCl\u2212][HOCl]=x\u00d7x0.452\u2212x\u2248x20.452K_a = \\frac{[H^+][OCl^-]}{[HOCl]} = \\frac{x \\times x}{0.452 – x} \\approx \\frac{x^2}{0.452}<\/p>\n\n\n\n Since xx is very small compared to 0.452, approximation holds.<\/p>\n\n\n\n Ka=(0.0001297)20.452=1.683\u00d710\u221280.452=3.724\u00d710\u22128K_a = \\frac{(0.0001297)^2}{0.452} = \\frac{1.683 \\times 10^{-8}}{0.452} = 3.724 \\times 10^{-8}<\/p>\n\n\n\n Ka\u22483.72\u00d710\u22128\\boxed{K_a \\approx 3.72 \\times 10^{-8}}<\/p>\n\n\n\n The acid dissociation constant KaK_a is a quantitative measure of the strength of a weak acid in solution. It defines the equilibrium concentration ratio of the dissociated ions to the undissociated acid. In this problem, we\u2019re dealing with hypochlorous acid (HOCl), a weak acid that partially ionizes in water to produce hydrogen ions (H+)(H^+) and hypochlorite ions (OCl\u2212)(OCl^-).<\/p>\n\n\n\n The problem provides the molarity of the acid solution and the percent ionization, which allows us to calculate the concentration of ions produced at equilibrium. The percent ionization is the fraction of the acid molecules that dissociate into ions relative to the initial acid concentration, multiplied by 100%. By converting the percent ionization into a decimal fraction, we find the equilibrium concentration of H+H^+ and OCl\u2212OCl^-.<\/p>\n\n\n\n Using an ICE table, we approximate the change in acid concentration as negligible because the ionization percentage is extremely small (less than 0.03%). This simplifies our calculations by allowing us to assume the initial acid concentration remains nearly the same at equilibrium.<\/p>\n\n\n\n The KaK_a is then calculated by the ratio of the product of the concentrations of the ionized species to the concentration of the undissociated acid. Since the concentrations of H+H^+ and OCl\u2212OCl^- are equal, KaK_a depends on the square of the ion concentration divided by the initial acid concentration.<\/p>\n\n\n\n The very small value of KaK_a confirms that hypochlorous acid is a weak acid, ionizing very slightly in solution, which is consistent with the low percent ionization observed experimentally.<\/p>\n","protected":false},"excerpt":{"rendered":" In the laboratory a student measures the percent ionization of a 0.452 M solution of hypochlorous acid to be 0.0287%. Calculate value of ka from this experimental data. The correct answer and explanation is: Let’s solve the problem step-by-step: Given: Step 1: Write the ionization reaction for hypochlorous acid (HOCl): HOCl\u21ccH++OCl\u2212\\mathrm{HOCl} \\rightleftharpoons \\mathrm{H^+} + \\mathrm{OCl^-} […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-23175","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23175","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=23175"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23175\/revisions"}],"predecessor-version":[{"id":23176,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/23175\/revisions\/23176"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=23175"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=23175"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=23175"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nGiven:<\/h3>\n\n\n\n
\n
\n\n\n\nStep 1: Write the ionization reaction for hypochlorous acid (HOCl):<\/h3>\n\n\n\n
\n\n\n\nStep 2: Calculate the concentration of ionized acid:<\/h3>\n\n\n\n
\n\n\n\nStep 3: Set up the ICE table (Initial, Change, Equilibrium):<\/h3>\n\n\n\n
Species<\/th> Initial (M)<\/th> Change (M)<\/th> Equilibrium (M)<\/th><\/tr><\/thead> HOCl<\/td> 0.452<\/td> \u2212x-x<\/td> 0.452\u2212x\u22480.4520.452 – x \\approx 0.452 (x is small)<\/td><\/tr> H+H^+<\/td> 0<\/td> +x+x<\/td> x=0.0001297x = 0.0001297<\/td><\/tr> OCl\u2212OCl^-<\/td> 0<\/td> +x+x<\/td> x=0.0001297x = 0.0001297<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\nStep 4: Write the expression for KaK_a:<\/h3>\n\n\n\n
\n\n\n\nStep 5: Calculate KaK_a:<\/h3>\n\n\n\n
\n\n\n\nFinal answer:<\/strong><\/h3>\n\n\n\n
\n\n\n\nExplanation (about 300 words):<\/h3>\n\n\n\n