To determine the ratio of hydrogen phosphate (HPO\u2084\u00b2\u207b) to dihydrogen phosphate (H\u2082PO\u2084\u207b) needed to create a buffer of pH 7.5, we use the Henderson-Hasselbalch equation<\/strong>:pH=pKa+log\u2061([base][acid])\\text{pH} = \\text{pKa} + \\log \\left( \\frac{[\\text{base}]}{[\\text{acid}]} \\right)pH=pKa+log([acid][base]\u200b)<\/p>\n\n\n\n In this reaction:<\/p>\n\n\n\n Given:<\/p>\n\n\n\n 7.5=6.8+log\u2061([HPO\u20842\u2212][H\u2082PO\u2084\u2212])7.5 = 6.8 + \\log \\left( \\frac{[\\text{HPO\u2084}^{2-}]}{[\\text{H\u2082PO\u2084}^-]} \\right)7.5=6.8+log([H\u2082PO\u2084\u2212][HPO\u20842\u2212]\u200b)<\/p>\n\n\n\n 7.5\u22126.8=log\u2061([HPO\u20842\u2212][H\u2082PO\u2084\u2212])7.5 – 6.8 = \\log \\left( \\frac{[\\text{HPO\u2084}^{2-}]}{[\\text{H\u2082PO\u2084}^-]} \\right)7.5\u22126.8=log([H\u2082PO\u2084\u2212][HPO\u20842\u2212]\u200b)0.7=log\u2061([HPO\u20842\u2212][H\u2082PO\u2084\u2212])0.7 = \\log \\left( \\frac{[\\text{HPO\u2084}^{2-}]}{[\\text{H\u2082PO\u2084}^-]} \\right)0.7=log([H\u2082PO\u2084\u2212][HPO\u20842\u2212]\u200b)<\/p>\n\n\n\n [HPO\u20842\u2212][H\u2082PO\u2084\u2212]=100.7\u22485.01\\frac{[\\text{HPO\u2084}^{2-}]}{[\\text{H\u2082PO\u2084}^-]} = 10^{0.7} \\approx 5.01[H\u2082PO\u2084\u2212][HPO\u20842\u2212]\u200b=100.7\u22485.01<\/p>\n\n\n\n The ratio of HPO\u2084\u00b2\u207b to H\u2082PO\u2084\u207b must be approximately 5.0 to 1<\/strong> in order to prepare a buffer solution at pH 7.5 using sodium dihydrogen phosphate.<\/p>\n\n\n\n A buffer solution is a mixture of a weak acid and its conjugate base, designed to resist changes in pH. The effectiveness of a buffer depends on the ratio of the concentrations of the base to the acid, and this is described using the Henderson-Hasselbalch equation.<\/p>\n\n\n\n In this case, we are given that sodium dihydrogen phosphate (NaH\u2082PO\u2084) has a pKa of 6.8. The conjugate base of this acid is hydrogen phosphate (HPO\u2084\u00b2\u207b). We want to create a buffer with a pH of 7.5. Since 7.5 is higher than 6.8, we will need more base than acid in our solution.<\/p>\n\n\n\n By rearranging and solving the Henderson-Hasselbalch equation, we find that the logarithmic difference between the desired pH and the pKa is 0.7. Taking the antilog (10 raised to the power of 0.7), we calculate the necessary ratio of base to acid to be about 5.0. This means we need five times as much hydrogen phosphate as dihydrogen phosphate to reach a pH of 7.5.<\/p>\n\n\n\n This ratio allows the buffer to maintain its pH when small amounts of acid or base are added, which is useful in biological and chemical systems where stable pH is essential.<\/p>\n\n\n\n The pKa of sodium dihydrogen phosphate (NaH2PO4) is 6.8. What ratio of hydrogen phosphate (HPO4^2-) to dihydrogen phosphate (H2PO4^-) would create a buffer of pH 7.5? The acid dissociation reaction of dihydrogen phosphate into hydrogen phosphate is shown below: H2PO4^- + H2O \u00e2\u2021\u0152 HPO4^2- + H3O+. pKa = 6.8 The Correct Answer and Explanation is: […]<\/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-39085","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39085","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=39085"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39085\/revisions"}],"predecessor-version":[{"id":39087,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/39085\/revisions\/39087"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=39085"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=39085"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=39085"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
\n
Step 1: Plug values into the equation<\/h3>\n\n\n\n
Step 2: Subtract pKa from both sides<\/h3>\n\n\n\n
Step 3: Take the antilogarithm (base 10)<\/h3>\n\n\n\n
Final Answer:<\/h3>\n\n\n\n
\n\n\n\nExplanation (around 300 words):<\/h3>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"