{"id":37769,"date":"2025-06-26T05:49:23","date_gmt":"2025-06-26T05:49:23","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=37769"},"modified":"2025-06-26T05:49:25","modified_gmt":"2025-06-26T05:49:25","slug":"sodium-hydroxide-naoh-reacts-with-sulfuric-acid-h2so4-according-to-the-equation","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/sodium-hydroxide-naoh-reacts-with-sulfuric-acid-h2so4-according-to-the-equation\/","title":{"rendered":"Sodium hydroxide (NaOH) reacts with sulfuric acid (H2SO4) according to the equation"},"content":{"rendered":"\n
Sodium hydroxide (NaOH) reacts with sulfuric acid (H2SO4) according to the equation: NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + 2H2O(l). In an experiment to make some sodium sulfate (Na2SO4), a chemist neutralizes 250 cm3 of 0.5 M (0.5 mol dm-3) H2SO4 with solid NaOH. The resulting solution is evaporated to dryness to recover the Na2SO4. Atomic masses: Na = 23, H = 1, S = 32, O = 16. – How many moles of H2SO4 are in 250 cm3 of 0.05 M sulfuric acid? – How many moles of NaOH does the chemist need to react with this amount of H2SO4? – What is the mass of NaOH required? – What is the number of moles of Na2SO4 produced? – What is the concentration of the Na2SO4 solution when the reaction is complete? – What mass of Na2SO4 would be produced after drying the product (assume 100% yield)? – A small experimental point: Why would the chemist need to be very careful in carrying out this reaction?<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
Correct Answers:<\/h3>\n\n\n\n\n
Moles of H\u2082SO\u2084 in 250 cm\u00b3 of 0.5 M solution<\/strong>: Volume = 250 cm\u00b3 = 0.250 dm\u00b3 Moles = Molarity \u00d7 Volume = 0.5 mol\/dm\u00b3 \u00d7 0.250 dm\u00b3 = 0.125 mol<\/strong><\/li>\n\n\n\n
Moles of NaOH needed<\/strong>: Balanced equation: 2NaOH + H\u2082SO\u2084 \u2192 Na\u2082SO\u2084 + 2H\u2082O From the ratio, 1 mole of H\u2082SO\u2084 reacts with 2 moles of NaOH So, moles of NaOH = 2 \u00d7 0.125 = 0.25 mol<\/strong><\/li>\n\n\n\n
Mass of NaOH required<\/strong>: Molar mass of NaOH = 23 + 16 + 1 = 40 g\/mol Mass = moles \u00d7 molar mass = 0.25 \u00d7 40 = 10 grams<\/strong><\/li>\n\n\n\n
Moles of Na\u2082SO\u2084 produced<\/strong>: From the equation, 1 mole H\u2082SO\u2084 produces 1 mole Na\u2082SO\u2084 So, moles of Na\u2082SO\u2084 = 0.125 mol<\/li>\n\n\n\n
Mass of Na\u2082SO\u2084 produced after drying<\/strong>: Molar mass of Na\u2082SO\u2084 = (2 \u00d7 23) + 32 + (4 \u00d7 16) = 46 + 32 + 64 = 142 g\/mol Mass = 0.125 mol \u00d7 142 g\/mol = 17.75 grams<\/strong><\/li>\n\n\n\n
Why the chemist needs to be careful<\/strong>: The reaction between NaOH and H\u2082SO\u2084 is exothermic. This means it releases heat. If the solid NaOH is added too quickly to the acid, the solution can become hot rapidly and boil or splash, which may be dangerous. The chemist should add the solid slowly with stirring to control the reaction and avoid overheating. Proper lab safety such as goggles and gloves should be used to prevent burns from hot acid or alkali.<\/li>\n<\/ol>\n\n\n\n\n\n\n\n
Explanation <\/h3>\n\n\n\n
This reaction is a classic acid\u2013base neutralization between sulfuric acid (H\u2082SO\u2084) and sodium hydroxide (NaOH), forming a salt (sodium sulfate) and water. To determine the correct quantities for this reaction, we start by converting the volume of the acid solution into liters, since molarity (mol\/dm\u00b3) relates to liters. For a 0.5 M solution and 0.250 dm\u00b3 volume, the moles of acid are 0.125.<\/p>\n\n\n\n
The balanced chemical equation shows a 1:2 molar ratio between H\u2082SO\u2084 and NaOH, meaning two moles of NaOH are needed per mole of acid. Therefore, 0.25 moles of NaOH are required to fully react with 0.125 moles of H\u2082SO\u2084.<\/p>\n\n\n\n
To find the mass of NaOH, we use its molar mass, which is the sum of sodium, oxygen, and hydrogen atomic masses. With 0.25 moles needed, we calculate a required mass of 10 grams.<\/p>\n\n\n\n
Since each mole of H\u2082SO\u2084 produces one mole of Na\u2082SO\u2084, the amount of product formed is 0.125 moles. The concentration of the final sodium sulfate solution remains 0.5 mol\/dm\u00b3 before evaporation, as no extra volume is added. Once the solution is evaporated, solid Na\u2082SO\u2084 remains. Using its molar mass, the dry mass of sodium sulfate is found to be 17.75 grams.<\/p>\n\n\n\n
Safety during this reaction is vital because it is exothermic. If NaOH is added too quickly, the heat can cause splattering or burns. Handling chemicals cautiously and using personal protective equipment ensures the procedure remains safe and controlled.<\/p>\n\n\n\n<\/figure>\n","protected":false},"excerpt":{"rendered":"
Sodium hydroxide (NaOH) reacts with sulfuric acid (H2SO4) according to the equation: NaOH(aq) + H2SO4(aq) -> Na2SO4(aq) + 2H2O(l). In an experiment to make some sodium sulfate (Na2SO4), a chemist neutralizes 250 cm3 of 0.5 M (0.5 mol dm-3) H2SO4 with solid NaOH. The resulting solution is evaporated to dryness to recover the Na2SO4. Atomic […]<\/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-37769","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37769","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=37769"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37769\/revisions"}],"predecessor-version":[{"id":37771,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37769\/revisions\/37771"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=37769"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=37769"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=37769"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
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