Methylamine, CH3NH2, is a weak base that hydrolyzes water according to the equation shown below. The value of ionization constant, Kb, is 5.2 x 10-4 (a) Calculate the concentration of OH in a 0.225M methylamine solution. (b) Calculate the pH.<\/p>\n\n\n\n
Methylamine, CH3<\/sub>NH2<\/sub>, is a weak base that hydrolyzes water according to the equation shown below. The value of ionization constant, Kb, is 5.2 x 10-4<\/sup><\/p>\n\n\n\n (a) Calculate the concentration of OH in a 0.225M methylamine solution.<\/p>\n\n\n\n (b) Calculate the pH.<\/p>\n\n\n\n The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s break this down step by step.<\/p>\n\n\n\n To solve for the concentration of OH\u207b, we will first set up an ICE (Initial, Change, Equilibrium) table.<\/p>\n\n\n\n Let\u2019s assume that the initial concentration of (CH_3NH_2) is 0.225 M and that none of the (CH_3NH_3^+) and OH\u207b have been formed at the start, so their initial concentrations are 0.<\/p>\n\n\n\n The expression for (K_b) is: Substituting the equilibrium concentrations into this equation: Thus, the equation simplifies to: To calculate the pH, we need to find the pOH first using the concentration of OH\u207b. The relationship between pOH and OH\u207b concentration is given by: Finally, we can calculate the pH using the relationship between pH and pOH: The weak base methylamine ((CH_3NH_2)) reacts with water to produce hydroxide ions (OH\u207b) and methylammonium ions ((CH_3NH_3^+)) through hydrolysis. The concentration of OH\u207b is determined by applying the ionization constant (K_b) in an equilibrium expression. Since (K_b) is small, we assume that the change in concentration is negligible compared to the initial concentration, simplifying the calculations. The concentration of OH\u207b is then used to determine the pOH, and from there, the pH is calculated. The pH value indicates a basic solution, which is consistent with the behavior of methylamine as a weak base.<\/p>\n","protected":false},"excerpt":{"rendered":" Methylamine, CH3NH2, is a weak base that hydrolyzes water according to the equation shown below. The value of ionization constant, Kb, is 5.2 x 10-4 (a) Calculate the concentration of OH in a 0.225M methylamine solution. (b) Calculate the pH. Methylamine, CH3NH2, is a weak base that hydrolyzes water according to the equation shown below. […]<\/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-191130","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191130","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=191130"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191130\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191130"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191130"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191130"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given Data:<\/h3>\n\n\n\n
\n
[
CH_3NH_2 + H_2O \\rightleftharpoons CH_3NH_3^+ + OH^-
]<\/li>\n<\/ul>\n\n\n\n(a) Calculation of OH\u207b Concentration<\/h3>\n\n\n\n
Species<\/th> Initial (M)<\/th> Change (M)<\/th> Equilibrium (M)<\/th><\/tr><\/thead> (CH_3NH_2)<\/td> 0.225<\/td> -x<\/td> 0.225 – x<\/td><\/tr> (CH_3NH_3^+)<\/td> 0<\/td> +x<\/td> x<\/td><\/tr> (OH^-)<\/td> 0<\/td> +x<\/td> x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
[
K_b = \\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}
]<\/p>\n\n\n\n
[
5.2 \\times 10^{-4} = \\frac{x \\cdot x}{0.225 – x}
]
Since (K_b) is small, we can assume that (x) is much smaller than 0.225 M, so we approximate (0.225 – x \\approx 0.225).<\/p>\n\n\n\n
[
5.2 \\times 10^{-4} = \\frac{x^2}{0.225}
]
Solving for (x):
[
x^2 = (5.2 \\times 10^{-4}) \\times 0.225 = 1.17 \\times 10^{-4}
]
[
x = \\sqrt{1.17 \\times 10^{-4}} \\approx 0.0108 \\, \\text{M}
]
So, the concentration of OH\u207b is approximately (0.0108 \\, \\text{M}).<\/p>\n\n\n\n(b) Calculation of pH<\/h3>\n\n\n\n
[
pOH = -\\log[OH^-]
]
Substitute the value of ( [OH^-] = 0.0108 \\, \\text{M} ):
[
pOH = -\\log(0.0108) \\approx 1.97
]<\/p>\n\n\n\n
[
pH + pOH = 14
]
[
pH = 14 – pOH = 14 – 1.97 = 12.03
]<\/p>\n\n\n\nFinal Answers:<\/h3>\n\n\n\n
\n
Explanation:<\/h3>\n\n\n\n