{"id":43931,"date":"2025-06-30T11:13:42","date_gmt":"2025-06-30T11:13:42","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=43931"},"modified":"2025-06-30T11:13:43","modified_gmt":"2025-06-30T11:13:43","slug":"the-ph-of-an-h2so4-solution-is-2","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-ph-of-an-h2so4-solution-is-2\/","title":{"rendered":"The pH of an H2SO4 solution is 2"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">The pH of an H2SO4 solution is 2. The pH of this solution, if an equal volume of 0.2 M NaCl solution is added to it, will be.<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><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 class=\"wp-block-paragraph\">When 0.2 M NaCl is added to a sulfuric acid (H\u2082SO\u2084) solution, it does not affect the pH significantly because NaCl is a neutral salt that does not undergo hydrolysis in water. Sodium chloride dissociates completely into Na\u207a and Cl\u207b ions, neither of which affects the concentration of H\u207a ions or the acidity of the solution.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-step breakdown:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Initial pH of H\u2082SO\u2084 Solution:<\/strong><br>The pH of the original sulfuric acid solution is 2. Sulfuric acid is a strong acid, and it dissociates completely in water, releasing two H\u207a ions per molecule.<ul><li>The concentration of H\u207a ions in a solution of H\u2082SO\u2084 can be calculated from the pH value using the formula:<\/li><\/ul>[H+]=10\u2212pH[\\text{H}^+] = 10^{-\\text{pH}}<ul><li>Since the pH is 2, the concentration of H\u207a ions is:<\/li><\/ul>[H+]=10\u22122=0.01\u2009M[\\text{H}^+] = 10^{-2} = 0.01 \\, \\text{M} This means that the H\u2082SO\u2084 solution has a concentration of 0.01 M H\u207a ions.<\/li>\n\n\n\n<li><strong>Effect of Adding NaCl Solution:<\/strong><br>When an equal volume of 0.2 M NaCl is added to the H\u2082SO\u2084 solution, the volume of the solution doubles, and the total moles of H\u207a ions remain the same. The H\u207a ions from H\u2082SO\u2084 will now be spread out in a larger volume, diluting the concentration of H\u207a ions.<ul><li>The new concentration of H\u207a ions after dilution is:<\/li><\/ul>[H+]=0.01\u2009M\u00d7VH\u2082SO\u20842VH\u2082SO\u2084=0.012=0.005\u2009M[\\text{H}^+] = \\frac{0.01 \\, \\text{M} \\times V_{\\text{H\u2082SO\u2084}}}{2V_{\\text{H\u2082SO\u2084}}} = \\frac{0.01}{2} = 0.005 \\, \\text{M}<ul><li>The pH can be calculated using the formula:<\/li><\/ul>pH=\u2212log\u2061[H+]\\text{pH} = -\\log[\\text{H}^+]<ul><li>So, the new pH is:<\/li><\/ul>pH=\u2212log\u2061(0.005)\u22482.3\\text{pH} = -\\log(0.005) \\approx 2.3<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">Final Result:<\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">The pH of the solution after adding an equal volume of 0.2 M NaCl will be approximately <strong>2.3<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The pH of an H2SO4 solution is 2. The pH of this solution, if an equal volume of 0.2 M NaCl solution is added to it, will be. The correct answer and explanation is: When 0.2 M NaCl is added to a sulfuric acid (H\u2082SO\u2084) solution, it does not affect the pH significantly because NaCl [&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-43931","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43931","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=43931"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43931\/revisions"}],"predecessor-version":[{"id":43932,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43931\/revisions\/43932"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=43931"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=43931"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=43931"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}