It takes 28.61 mL of 0.216 M sodium hydroxide solution to titrate 10.00 mL of sulfuric acid solution. What is the molarity of the sulfuric acid?<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To find the molarity of the sulfuric acid (H\u2082SO\u2084) solution, we can use the concept of molarity and the balanced chemical equation for the titration reaction between sodium hydroxide (NaOH) and sulfuric acid (H\u2082SO\u2084).<\/p>\n\n\n\n The balanced equation for the reaction between sodium hydroxide and sulfuric acid is:<\/p>\n\n\n\n $$ From this equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of sodium hydroxide.<\/p>\n\n\n\n The titration formula is:<\/p>\n\n\n\n $$ Where:<\/p>\n\n\n\n We are given:<\/p>\n\n\n\n According to the balanced equation, 2 moles of NaOH react with 1 mole of H\u2082SO\u2084, so we need to account for this ratio by dividing the volume of NaOH by 2 in the calculation.<\/p>\n\n\n\n Thus, the effective volume of NaOH is:<\/p>\n\n\n\n $$ Now, using the titration formula, we substitute the values and solve for $M_2$:<\/p>\n\n\n\n $$ $$ $$ $$ The molarity of the sulfuric acid solution is 0.309 M<\/strong>.<\/p>\n\n\n\n In this titration problem, we used the concept of molarity, the stoichiometric relationship between NaOH and H\u2082SO\u2084, and the titration formula to determine the molarity of the sulfuric acid solution. By adjusting the volume of NaOH for the 2:1 ratio with H\u2082SO\u2084 and applying the formula, we obtained the final result.<\/p>\n","protected":false},"excerpt":{"rendered":" It takes 28.61 mL of 0.216 M sodium hydroxide solution to titrate 10.00 mL of sulfuric acid solution. What is the molarity of the sulfuric acid? The correct answer and explanation is : To find the molarity of the sulfuric acid (H\u2082SO\u2084) solution, we can use the concept of molarity and the balanced chemical equation […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-1065","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/1065","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=1065"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/1065\/revisions"}],"predecessor-version":[{"id":1066,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/1065\/revisions\/1066"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=1065"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=1065"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=1065"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Write the balanced chemical equation<\/h3>\n\n\n\n
H_2SO_4 (aq) + 2NaOH (aq) \\rightarrow Na_2SO_4 (aq) + 2H_2O (l)
$$<\/p>\n\n\n\nStep 2: Use the titration formula<\/h3>\n\n\n\n
M_1 V_1 = M_2 V_2
$$<\/p>\n\n\n\n\n
\n
Step 3: Adjust for the stoichiometric ratio<\/h3>\n\n\n\n
V_1′ = \\frac{V_1}{2} = \\frac{0.02861}{2} = 0.014305 \\, \\text{L}
$$<\/p>\n\n\n\nStep 4: Solve for $M_2$, the molarity of sulfuric acid<\/h3>\n\n\n\n
M_1 V_1′ = M_2 V_2
$$<\/p>\n\n\n\n
0.216 \\, \\text{M} \\times 0.014305 \\, \\text{L} = M_2 \\times 0.01000 \\, \\text{L}
$$<\/p>\n\n\n\n
M_2 = \\frac{0.216 \\times 0.014305}{0.01000}
$$<\/p>\n\n\n\n
M_2 = 0.309 \\, \\text{M}
$$<\/p>\n\n\n\nFinal Answer:<\/h3>\n\n\n\n
Explanation:<\/h3>\n\n\n\n