Methylamine, CH3NH2, is a weak base that reacts according to the reaction
CH3NH2 + H2O <–> CH3NH3 + OH”
The value of the ionization constant, Kb, is 5.25 x 104. Methylamine reacts to form salts such as methylammonium nitrate, (CH3NH3)
(NO3).
b. Calculate the pH of a solution made by adding 0.0150 mole of solid methylammonium nitrate to 150.0 mL of a 0.125 molar solution of methylamine. Assume that no volume change occurs.
Answer:
Methylamine, CH3NH2, is a weak base that reacts according to the reaction
CH3NH2 + H2O <-> CH3NH3 + OH”
The value of the ionization constant, Kb, is 5.25 x 104. Methylamine reacts to form salts such as methylammonium nitrate, (CH3NH3″)
(NO3).
How many moles of either NaOH or HCI (state clearly which you choose) should be added to the solution in (b) to produce a solution with
a pH of 11.00? Assume that no volume change occurs.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n To calculate the pH of the solution made by adding methylammonium nitrate (CH\u2083NH\u2083NO\u2083) to the methylamine (CH\u2083NH\u2082) solution, we need to consider the reactions and equilibrium between methylamine and water, as well as the contribution of the methylammonium ion (CH\u2083NH\u2083\u207a), which acts as a weak acid.<\/p>\n\n\n\n The concentration of CH\u2083NH\u2083\u207a is initially 0.0150 moles in 0.150 L, so:<\/p>\n\n\n\n [ Now, for the dissociation of CH\u2083NH\u2083\u207a in water:<\/p>\n\n\n\n [ The change in concentration can be written as:<\/p>\n\n\n\n [ The equilibrium expression for Ka is:<\/p>\n\n\n\n [ Substitute the known value of Ka:<\/p>\n\n\n\n [ Assume ( x ) is small compared to 0.100 M (because Ka is small), so ( 0.100 – x \\approx 0.100 ):<\/p>\n\n\n\n [ Solving for ( x ):<\/p>\n\n\n\n [ [ Thus, the concentration of H\u2083O\u207a is ( 1.38 \\times 10^{-6} ) M. The pH is:<\/p>\n\n\n\n [ Therefore, the pH of the solution is approximately 5.86<\/strong>.<\/p>\n\n\n\n To achieve a pH of 11.00, we need to adjust the concentration of ( H\u2083O\u207a ) to a level that corresponds to this pH. The target pH of 11.00 corresponds to a target ( [H\u2083O\u207a] ) of:<\/p>\n\n\n\n [ Since the initial concentration of ( H\u2083O\u207a ) is ( 1.38 \\times 10^{-6} \\, \\text{M} ), we need to reduce the ( [H\u2083O\u207a] ) concentration. This can be done by adding NaOH (a strong base) to neutralize some of the ( H\u2083O\u207a ).<\/p>\n\n\n\n To calculate the moles of NaOH needed, we use the change in ( [H\u2083O\u207a] ):<\/p>\n\n\n\n [ Now, the moles of ( H\u2083O\u207a ) to be neutralized:<\/p>\n\n\n\n [ Since NaOH and H\u2083O\u207a react in a 1:1 ratio, the moles of NaOH required are the same:<\/p>\n\n\n\n [ Thus, to achieve a pH of 11.00, you should add approximately 2.07 \u00d7 10\u207b\u2077 moles of NaOH<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" Methylamine, CH3NH2, is a weak base that reacts according to the reactionCH3NH2 + H2O <–> CH3NH3 + OH”The value of the ionization constant, Kb, is 5.25 x 104. Methylamine reacts to form salts such as methylammonium nitrate, (CH3NH3)(NO3).b. Calculate the pH of a solution made by adding 0.0150 mole of solid methylammonium nitrate to 150.0 […]<\/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-191128","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191128","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=191128"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191128\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191128"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191128"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191128"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Part 1: Calculating the pH of the Solution<\/h3>\n\n\n\n
Step 1: Write the equilibrium equations and determine known quantities.<\/h4>\n\n\n\n
\n
CH\u2083NH\u2082 + H\u2082O \\rightleftharpoons CH\u2083NH\u2083\u207a + OH\u207b
]
Kb for CH\u2083NH\u2082 = 5.25 x 10\u207b\u2074.<\/li>\n\n\n\n
CH\u2083NH\u2083\u207a + H\u2082O \\rightleftharpoons CH\u2083NH\u2082 + H\u2083O\u207a
]
The ionization constant Ka for CH\u2083NH\u2083\u207a can be found using the relationship between Ka and Kb: [
K_a = \\frac{K_w}{K_b} = \\frac{1.0 \\times 10^{-14}}{5.25 \\times 10^{-4}} = 1.90 \\times 10^{-11}
]<\/li>\n<\/ul>\n\n\n\nStep 2: Set up the equilibrium for CH\u2083NH\u2083\u207a dissociation.<\/h4>\n\n\n\n
\\text{Concentration of CH\u2083NH\u2083\u207a} = \\frac{0.0150 \\, \\text{mol}}{0.150 \\, \\text{L}} = 0.100 \\, \\text{M}
]<\/p>\n\n\n\n
\\text{CH\u2083NH\u2083\u207a} + H\u2082O \\rightleftharpoons CH\u2083NH\u2082 + H\u2083O\u207a
]<\/p>\n\n\n\n
\\begin{aligned}
\\text{Initial concentration:} & \\quad [CH\u2083NH\u2083\u207a] = 0.100 \\, \\text{M} \\quad \\text{(CH\u2083NH\u2082 and H\u2083O\u207a are initially 0)} \\
\\text{Change in concentration:} & \\quad -x \\quad \\text{(for CH\u2083NH\u2083\u207a, +x for CH\u2083NH\u2082 and +x for H\u2083O\u207a)} \\
\\text{Equilibrium concentrations:} & \\quad [CH\u2083NH\u2083\u207a] = 0.100 – x, \\quad [CH\u2083NH\u2082] = x, \\quad [H\u2083O\u207a] = x
\\end{aligned}
]<\/p>\n\n\n\n
K_a = \\frac{[CH\u2083NH\u2082][H\u2083O\u207a]}{[CH\u2083NH\u2083\u207a]} = \\frac{x^2}{0.100 – x}
]<\/p>\n\n\n\n
1.90 \\times 10^{-11} = \\frac{x^2}{0.100 – x}
]<\/p>\n\n\n\n
1.90 \\times 10^{-11} = \\frac{x^2}{0.100}
]<\/p>\n\n\n\n
x^2 = (1.90 \\times 10^{-11}) \\times 0.100 = 1.90 \\times 10^{-12}
]<\/p>\n\n\n\n
x = \\sqrt{1.90 \\times 10^{-12}} = 1.38 \\times 10^{-6} \\, \\text{M}
]<\/p>\n\n\n\n
\\text{pH} = -\\log[H\u2083O\u207a] = -\\log(1.38 \\times 10^{-6}) = 5.86
]<\/p>\n\n\n\nPart 2: Adding NaOH or HCl to Achieve a pH of 11.00<\/h3>\n\n\n\n
\\text{pH} = 11.00 \\quad \\Rightarrow \\quad [H\u2083O\u207a] = 10^{-11} \\, \\text{M}
]<\/p>\n\n\n\n
\\Delta[H\u2083O\u207a] = (1.38 \\times 10^{-6} – 10^{-11}) \\, \\text{M} \\approx 1.38 \\times 10^{-6} \\, \\text{M}
]<\/p>\n\n\n\n
\\text{Moles of } H\u2083O\u207a = 1.38 \\times 10^{-6} \\, \\text{M} \\times 0.150 \\, \\text{L} = 2.07 \\times 10^{-7} \\, \\text{mol}
]<\/p>\n\n\n\n
\\text{Moles of NaOH} = 2.07 \\times 10^{-7} \\, \\text{mol}
]<\/p>\n\n\n\n