What is the Kb of a base that has a pH of 8.98 when in a 0.186M aqueous solution.<\/p>\n\n\n\n
The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n Correct Answer:<\/strong> To find the base dissociation constant (Kb<\/strong>) of a base in a 0.186 M solution with a pH of 8.98, we follow these steps:<\/p>\n\n\n\n Step 1: Find the pOH from pH<\/strong> Step 2: Find the hydroxide ion concentration [OH\u207b]<\/strong> Step 3: Set up an ICE table for the weak base (B)<\/strong> Initially:<\/p>\n\n\n\n Change:<\/p>\n\n\n\n At equilibrium:<\/p>\n\n\n\n Step 4: Use the Kb expression<\/strong> Kb=[BH+][OH\u2212][B]=(9.55\u00d710\u22126)20.186K_b = \\frac{[BH^+][OH^-]}{[B]} = \\frac{(9.55 \\times 10^{-6})^2}{0.186} Kb=9.12\u00d710\u2212110.186\u22484.9\u00d710\u221210K_b = \\frac{9.12 \\times 10^{-11}}{0.186} \\approx 4.9 \\times 10^{-10}<\/p>\n\n\n\n Correction:<\/strong> The square of 9.55 \u00d7 10\u207b\u2076 is actually closer to 9.12 \u00d7 10\u207b\u00b9\u00b9<\/strong>, and dividing that by 0.186 gives approximately 4.9 \u00d7 10\u207b\u00b9\u2070<\/strong>. However, there seems to be an error because the expected Kb for a pH of 8.98 in a 0.186 M solution is higher.<\/p>\n\n\n\n Let’s redo the math more carefully: (9.55\u00d710\u22126)2=9.12\u00d710\u221211(9.55 \\times 10^{-6})^2 = 9.12 \\times 10^{-11} Kb=9.12\u00d710\u2212110.186=4.9\u00d710\u221210K_b = \\frac{9.12 \\times 10^{-11}}{0.186} = 4.9 \\times 10^{-10}<\/p>\n\n\n\n This is accurate. Therefore, the final Kb \u2248 4.9 \u00d7 10\u207b\u00b9\u2070<\/strong>.<\/p>\n\n\n\n So, the correct answer is: What is the Kb of a base that has a pH of 8.98 when in a 0.186M aqueous solution. The correct answer and explanation is: Correct Answer:The base dissociation constant (Kb) is 5.6 \u00d7 10\u207b\u2076. Explanation (300 words) To find the base dissociation constant (Kb) of a base in a 0.186 M solution with a […]<\/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-39817","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39817","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=39817"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39817\/revisions"}],"predecessor-version":[{"id":39819,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39817\/revisions\/39819"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39817"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39817"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39817"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
The base dissociation constant (Kb<\/strong>) is 5.6 \u00d7 10\u207b\u2076<\/strong>.<\/p>\n\n\n\n
\n\n\n\nExplanation (300 words)<\/strong><\/h3>\n\n\n\n
Since pH + pOH = 14,
pOH = 14 – pH = 14 – 8.98 = 5.02<\/strong><\/p>\n\n\n\n
[OH\u207b]=10\u2212pOH=10\u22125.02\u22489.55\u00d710\u22126\u2009M\\text{[OH\u207b]} = 10^{-\\text{pOH}} = 10^{-5.02} \\approx 9.55 \\times 10^{-6} \\, \\text{M}<\/p>\n\n\n\n
Let B represent the base. Its reaction in water is:
B+H\u2082O\u21ccBH\u207a+OH\u207b\\text{B} + \\text{H\u2082O} \u21cc \\text{BH\u207a} + \\text{OH\u207b}<\/p>\n\n\n\n\n
\n
\n
Kb \u2248 4.9 \u00d7 10\u207b\u00b9\u2070<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"