a. Calculate percent ionization of .085 M lactic acid (Ka = 1.4 x 10-4).<\/p>\n\n\n\n
b. Calculate the percent ionization in a solution containing .085 M lactic acid and .050 M sodiumlactate.<\/p>\n\n\n\n
c. Explain the difference in percent ionization between the solutions in 2a and 2b.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the percent ionization of lactic acid ((\\text{C}_3\\text{H}_6\\text{O}_3)) in water, we can use the expression for the acid dissociation constant ((K_a)):<\/p>\n\n\n\n [ Given:<\/p>\n\n\n\n Let (x) be the concentration of ( \\text{H}^+ ) and ( \\text{C}_3\\text{H}_5\\text{O}_3^- ) at equilibrium. The equilibrium concentrations will be:<\/p>\n\n\n\n The equilibrium expression for the dissociation is:<\/p>\n\n\n\n [ Substitute the equilibrium concentrations:<\/p>\n\n\n\n [ Since (K_a) is small, we can assume that (x) will be much smaller than 0.085, so (0.085 – x \\approx 0.085). Thus, the equation simplifies to:<\/p>\n\n\n\n [ Solving for (x):<\/p>\n\n\n\n [ Now, we can calculate the percent ionization:<\/p>\n\n\n\n [ In a solution containing both lactic acid and its conjugate base (sodium lactate), the acid dissociation will be influenced by the common ion effect. Sodium lactate dissociates completely into ( \\text{Na}^+ ) and ( \\text{C}_3\\text{H}_5\\text{O}_3^- ), so the concentration of ( \\text{C}_3\\text{H}_5\\text{O}_3^- ) is 0.050 M from the salt.<\/p>\n\n\n\n This addition of the conjugate base decreases the ionization of lactic acid. The equilibrium expression is:<\/p>\n\n\n\n [ Let (x) represent the change in the concentration of ( \\text{H}^+ ) (and also ( \\text{C}_3\\text{H}_5\\text{O}_3^- )). At equilibrium:<\/p>\n\n\n\n Now, substitute into the expression for (K_a):<\/p>\n\n\n\n [ Assume that (x) will be small compared to 0.050 and 0.085, so we approximate the denominator as 0.085 and the numerator as (0.050x). This simplifies to:<\/p>\n\n\n\n [ Solve for (x):<\/p>\n\n\n\n [ Now, calculate the percent ionization:<\/p>\n\n\n\n [ The percent ionization of lactic acid is much higher in the absence of sodium lactate (4.06%) than in the solution containing sodium lactate (0.28%). This difference is due to the common ion effect<\/strong>, where the addition of the conjugate base ( \\text{C}_3\\text{H}_5\\text{O}_3^- ) from sodium lactate shifts the equilibrium of the dissociation of lactic acid to the left.<\/p>\n\n\n\n In the solution without sodium lactate, there are no other sources of ( \\text{C}_3\\text{H}_5\\text{O}_3^- ), so lactic acid dissociates more, resulting in higher ionization. However, in the presence of sodium lactate, the ( \\text{C}_3\\text{H}_5\\text{O}_3^- ) ions from the salt increase the concentration of the conjugate base, which suppresses the dissociation of lactic acid, leading to a lower percent ionization. This effect demonstrates Le Chatelier’s Principle, where the system tries to counteract the increase in the concentration of the conjugate base by reducing the dissociation of the acid.<\/p>\n","protected":false},"excerpt":{"rendered":" a. Calculate percent ionization of .085 M lactic acid (Ka = 1.4 x 10-4). b. Calculate the percent ionization in a solution containing .085 M lactic acid and .050 M sodiumlactate. c. Explain the difference in percent ionization between the solutions in 2a and 2b. The correct answer and explanation is : a. Calculation of […]<\/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-201205","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201205","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=201205"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201205\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201205"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201205"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201205"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}a. Calculation of Percent Ionization of 0.085 M Lactic Acid<\/h3>\n\n\n\n
\\text{C}_3\\text{H}_6\\text{O}_3 (aq) \\rightleftharpoons \\text{H}^+ (aq) + \\text{C}_3\\text{H}_5\\text{O}_3^- (aq)
]<\/p>\n\n\n\n\n
\n
K_a = \\frac{[\\text{H}^+][\\text{C}_3\\text{H}_5\\text{O}_3^-]}{[\\text{C}_3\\text{H}_6\\text{O}_3]}
]<\/p>\n\n\n\n
1.4 \\times 10^{-4} = \\frac{x^2}{0.085 – x}
]<\/p>\n\n\n\n
1.4 \\times 10^{-4} = \\frac{x^2}{0.085}
]<\/p>\n\n\n\n
x^2 = (1.4 \\times 10^{-4})(0.085)
]
[
x^2 = 1.19 \\times 10^{-5}
]
[
x = 3.45 \\times 10^{-3} \\, \\text{M}
]<\/p>\n\n\n\n
\\text{Percent ionization} = \\frac{x}{[\\text{HA}]_0} \\times 100 = \\frac{3.45 \\times 10^{-3}}{0.085} \\times 100 \\approx 4.06\\%
]<\/p>\n\n\n\nb. Calculation of Percent Ionization in a Solution Containing 0.085 M Lactic Acid and 0.050 M Sodium Lactate<\/h3>\n\n\n\n
K_a = \\frac{[\\text{H}^+][\\text{C}_3\\text{H}_5\\text{O}_3^-]}{[\\text{C}_3\\text{H}_6\\text{O}_3]}
]<\/p>\n\n\n\n\n
1.4 \\times 10^{-4} = \\frac{x(0.050 + x)}{0.085 – x}
]<\/p>\n\n\n\n
1.4 \\times 10^{-4} = \\frac{0.050x}{0.085}
]<\/p>\n\n\n\n
x = \\frac{1.4 \\times 10^{-4} \\times 0.085}{0.050} = 2.36 \\times 10^{-4} \\, \\text{M}
]<\/p>\n\n\n\n
\\text{Percent ionization} = \\frac{x}{[\\text{HA}]_0} \\times 100 = \\frac{2.36 \\times 10^{-4}}{0.085} \\times 100 \\approx 0.28\\%
]<\/p>\n\n\n\nc. Explanation of the Difference in Percent Ionization<\/h3>\n\n\n\n