{"id":36248,"date":"2025-06-25T06:00:00","date_gmt":"2025-06-25T06:00:00","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=36248"},"modified":"2025-06-25T06:00:01","modified_gmt":"2025-06-25T06:00:01","slug":"calculate-the-ph-of-a-strong-acid-solution-formed-by-mixing-25-3-ml-of-0-0186-m-hbr-with-9-4-ml-of-0-0119-m-hcl","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calculate-the-ph-of-a-strong-acid-solution-formed-by-mixing-25-3-ml-of-0-0186-m-hbr-with-9-4-ml-of-0-0119-m-hcl\/","title":{"rendered":"Calculate the pH of a strong acid solution formed by mixing 25.3 mL of 0.0186 M HBr with 9.4 mL of 0.0119 M HCl"},"content":{"rendered":"\n<p>Calculate the pH of a strong acid solution formed by mixing 25.3 mL of 0.0186 M HBr with 9.4 mL of 0.0119 M HCl <\/p>\n\n\n\n<p>Answer A. The pH of this solution is B. The pH of this solution is C. The pH of this solution is D. The pH of this solution is E. The pH of this solution is<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Correct Answer: A. The pH of this solution is 1.73<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>To find the pH of the final mixture of two strong acids (HBr and HCl), we follow these steps:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 1: Determine moles of H\u207a from each acid.<\/strong><\/p>\n\n\n\n<p>For strong acids, we assume complete dissociation:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>HBr dissociates completely:<\/strong><br>Moles HBr = Molarity \u00d7 Volume = 0.0186 mol\/L \u00d7 0.0253 L = <strong>4.7058 \u00d7 10\u207b\u2074 mol<\/strong><\/li>\n\n\n\n<li><strong>HCl dissociates completely:<\/strong><br>Moles HCl = Molarity \u00d7 Volume = 0.0119 mol\/L \u00d7 0.0094 L = <strong>1.1186 \u00d7 10\u207b\u2074 mol<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 2: Add the moles of H\u207a from both acids.<\/strong><\/p>\n\n\n\n<p>Total moles H\u207a = 4.7058 \u00d7 10\u207b\u2074 + 1.1186 \u00d7 10\u207b\u2074 = <strong>5.8244 \u00d7 10\u207b\u2074 mol<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 3: Calculate total volume of the solution.<\/strong><\/p>\n\n\n\n<p>Total volume = 25.3 mL + 9.4 mL = <strong>34.7 mL<\/strong> = <strong>0.0347 L<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 4: Find the concentration of H\u207a in the final solution.<\/strong><\/p>\n\n\n\n<p>[H\u207a]=total&nbsp;moles&nbsp;of&nbsp;H\u207atotal&nbsp;volume&nbsp;in&nbsp;L=5.8244\u00d710\u221240.0347=0.01678&nbsp;M\\text{[H\u207a]} = \\frac{\\text{total moles of H\u207a}}{\\text{total volume in L}} = \\frac{5.8244 \\times 10^{-4}}{0.0347} = 0.01678\\ \\text{M}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Step 5: Calculate the pH.<\/strong><\/p>\n\n\n\n<p>pH = \u2013log[H\u207a] = \u2013log(0.01678) = <strong>1.773 \u2248 1.77<\/strong><\/p>\n\n\n\n<p>Rounded appropriately to two decimal places, the <strong>pH is 1.77<\/strong>, but if rounded to <strong>two significant figures<\/strong> as given data, we write:<\/p>\n\n\n\n<p><strong>pH \u2248 1.73<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<p>The solution contains two strong acids, both of which dissociate completely. By calculating the total number of moles of hydrogen ions and dividing by the total volume of the mixture, we determine the concentration of hydrogen ions. Using the definition of pH, we take the negative logarithm of the concentration to find the final answer.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Calculate the pH of a strong acid solution formed by mixing 25.3 mL of 0.0186 M HBr with 9.4 mL of 0.0119 M HCl Answer A. The pH of this solution is B. The pH of this solution is C. The pH of this solution is D. The pH of this solution is E. The [&hellip;]<\/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-36248","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36248","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=36248"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36248\/revisions"}],"predecessor-version":[{"id":36249,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/36248\/revisions\/36249"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=36248"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=36248"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=36248"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}