What is the pH of a 0.56 M solution of dimethylamine?<\/p>\n\n\n\n
The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n To determine the pH of a 0.56 M solution of dimethylamine (DMA), we need to understand its nature as a weak base and apply the equilibrium expression for its dissociation in water.<\/p>\n\n\n\n Dimethylamine is a weak base and will react with water in the following way: DMA+H2O\u21ccDMAH++OH\u2212\\text{DMA} + \\text{H}_2\\text{O} \\rightleftharpoons \\text{DMAH}^+ + \\text{OH}^-<\/p>\n\n\n\n Here, DMA (dimethylamine) accepts a proton from water, forming DMAH+ (dimethylammonium ion) and hydroxide ions (OH-).<\/p>\n\n\n\n To calculate the pH, we first need the base dissociation constant (Kb) for dimethylamine. The Kb for dimethylamine is typically given as: Kb=5.4\u00d710\u22124\\text{Kb} = 5.4 \\times 10^{-4}<\/p>\n\n\n\n Using the formula for Kb: Kb=[DMAH+][OH\u2212][DMA]\\text{Kb} = \\frac{[\\text{DMAH}^+][\\text{OH}^-]}{[\\text{DMA}]}<\/p>\n\n\n\n At equilibrium, the concentration of DMAH+ and OH- will be equal, and the concentration of DMA will decrease by this amount.<\/p>\n\n\n\n Kb=x20.56\u2212x\\text{Kb} = \\frac{x^2}{0.56 – x}<\/p>\n\n\n\n Since the Kb value is small, we can assume that xx is much smaller than 0.56, so we approximate 0.56\u2212x\u22480.560.56 – x \\approx 0.56. 5.4\u00d710\u22124=x20.565.4 \\times 10^{-4} = \\frac{x^2}{0.56}<\/p>\n\n\n\n Solving for xx (which represents [OH-]): x2=(5.4\u00d710\u22124)(0.56)=3.024\u00d710\u22124x^2 = (5.4 \\times 10^{-4})(0.56) = 3.024 \\times 10^{-4} x=3.024\u00d710\u22124=0.0174\u2009Mx = \\sqrt{3.024 \\times 10^{-4}} = 0.0174 \\, \\text{M}<\/p>\n\n\n\n So, the concentration of OH- is approximately 0.0174 M.<\/p>\n\n\n\n The pOH is given by: pOH=\u2212log\u2061[OH\u2212]\\text{pOH} = -\\log [\\text{OH}^-]<\/p>\n\n\n\n Substitute the value of [OH-]: pOH=\u2212log\u2061(0.0174)=1.76\\text{pOH} = -\\log(0.0174) = 1.76<\/p>\n\n\n\n Finally, the pH can be found using the relationship: pH=14\u2212pOH\\text{pH} = 14 – \\text{pOH} pH=14\u22121.76=12.24\\text{pH} = 14 – 1.76 = 12.24<\/p>\n\n\n\n Thus, the pH of the 0.56 M solution of dimethylamine is approximately 12.24<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" What is the pH of a 0.56 M solution of dimethylamine? The correct answer and explanation is: To determine the pH of a 0.56 M solution of dimethylamine (DMA), we need to understand its nature as a weak base and apply the equilibrium expression for its dissociation in water. Dimethylamine is a weak base and […]<\/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-43779","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43779","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=43779"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43779\/revisions"}],"predecessor-version":[{"id":43780,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43779\/revisions\/43780"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=43779"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=43779"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=43779"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Set up the equilibrium expression<\/h3>\n\n\n\n
Step 2: Set up the ICE table<\/h3>\n\n\n\n
Species<\/th> Initial Concentration<\/th> Change in Concentration<\/th> Equilibrium Concentration<\/th><\/tr><\/thead> DMA<\/td> 0.56 M<\/td> -x<\/td> 0.56 – x<\/td><\/tr> DMAH+<\/td> 0<\/td> +x<\/td> x<\/td><\/tr> OH-<\/td> 0<\/td> +x<\/td> x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n Step 3: Substitute into the Kb expression<\/h3>\n\n\n\n
Step 4: Calculate the pOH<\/h3>\n\n\n\n
Step 5: Calculate the pH<\/h3>\n\n\n\n