Use the Henderson\u2013Hasselbalch equation to calculate the pKa of the acid.<\/p>\n\n\n\n
The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n Okay, to use the Henderson-Hasselbalch equation to calculate the pKa, we need values for the pH of a solution containing the acid and its conjugate base, as well as the concentrations of the acid and its conjugate base. Since no specific values were provided in the prompt, I will create a common scenario using typical concentrations to demonstrate the calculation.<\/p>\n\n\n\n Scenario:<\/strong> Assume we have a buffer solution prepared from a weak acid (HA) and its conjugate base (A\u207b). We measure the pH of this solution to be 4.50. We also know the concentration of the weak acid [HA] in the solution is 0.10 M, and the concentration of the conjugate base [A\u207b] is 0.20 M.<\/p>\n\n\n\n Correct Answer (Calculated pKa):<\/strong><\/p>\n\n\n\n Using the values from the scenario: The Henderson-Hasselbalch equation is: To find pKa, we rearrange the equation: Substitute the values: Rounding to two decimal places, the calculated pKa is approximately 4.20<\/strong>.<\/p>\n\n\n\n Explanation (Approximately 300 words):<\/strong><\/p>\n\n\n\n The Henderson-Hasselbalch equation is a fundamental relationship in chemistry used to calculate the pH of a buffer solution or, as requested here, to determine the pKa of a weak acid. The equation is expressed as:<\/p>\n\n\n\n pH = pKa + log ([A\u207b] \/ [HA])<\/strong><\/p>\n\n\n\n Here, pH represents the measured acidity of the solution, pKa is the negative logarithm of the acid dissociation constant (Ka) which is a characteristic property of the specific acid, [A\u207b] is the molar concentration of the conjugate base, and [HA] is the molar concentration of the weak acid.<\/p>\n\n\n\n To calculate the pKa of an acid using this equation, we must rearrange it to isolate pKa:<\/p>\n\n\n\n pKa = pH – log ([A\u207b] \/ [HA])<\/strong><\/p>\n\n\n\n This rearrangement shows that if we know the pH of a solution containing a significant amount of both the weak acid and its conjugate base (i.e., a buffer solution), and we know their respective molar concentrations, we can directly calculate the pKa.<\/p>\n\n\n\n Consider the scenario above: we had a solution with a measured pH of 4.50, an acid concentration [HA] of 0.10 M, and a conjugate base concentration [A\u207b] of 0.20 M. By plugging these values into the rearranged Henderson-Hasselbalch equation:<\/p>\n\n\n\n pKa = 4.50 – log (0.20 \/ 0.10) Calculating the logarithm of 2.0 gives approximately 0.301. Subtracting this from the pH:<\/p>\n\n\n\n pKa = 4.50 – 0.301 \u2248 4.199<\/p>\n\n\n\n Therefore, based on these measurements, the calculated pKa of the acid is approximately 4.20. The pKa value is a constant specific to the acid at a given temperature and provides a quantitative measure of its strength \u2013 a lower pKa indicates a stronger acid. The Henderson-Hasselbalch equation is most accurate when used in the buffer region where the ratio of [A\u207b]\/[HA] is close to 1, and the concentrations are not extremely dilute or concentrated.<\/p>\n","protected":false},"excerpt":{"rendered":" Use the Henderson\u2013Hasselbalch equation to calculate the pKa of the acid. The correct answer and explanation is: Okay, to use the Henderson-Hasselbalch equation to calculate the pKa, we need values for the pH of a solution containing the acid and its conjugate base, as well as the concentrations of the acid and its conjugate base. […]<\/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-15678","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/15678","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=15678"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/15678\/revisions"}],"predecessor-version":[{"id":15679,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/15678\/revisions\/15679"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=15678"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=15678"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=15678"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
pH = 4.50
[HA] = 0.10 M
[A\u207b] = 0.20 M<\/p>\n\n\n\n
pH = pKa + log ([A\u207b] \/ [HA])<\/p>\n\n\n\n
pKa = pH – log ([A\u207b] \/ [HA])<\/p>\n\n\n\n
pKa = 4.50 – log (0.20 M \/ 0.10 M)
pKa = 4.50 – log (2.0)
pKa \u2248 4.50 – 0.301
pKa \u2248 4.199<\/p>\n\n\n\n
\n\n\n\n
pKa = 4.50 – log (2.0)<\/p>\n\n\n\n