{"id":41492,"date":"2025-06-28T08:57:59","date_gmt":"2025-06-28T08:57:59","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=41492"},"modified":"2025-06-28T08:58:00","modified_gmt":"2025-06-28T08:58:00","slug":"calcium-sulfate-dihydrate-caso4-ae%c2%a2-2h2o-has-a-molar-mass-of-172-17-g-mol","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/calcium-sulfate-dihydrate-caso4-ae%c2%a2-2h2o-has-a-molar-mass-of-172-17-g-mol\/","title":{"rendered":"Calcium sulfate dihydrate (CaSO4 \u00e2\u20ac\u00a2 2H2O) has a molar mass of 172.17 g\/mol."},"content":{"rendered":"\n

Calcium sulfate dihydrate (CaSO4 \u00e2\u20ac\u00a2 2H2O) has a molar mass of 172.17 g\/mol. Considering that calcium sulfate dihydrate is a salt, what would be the pH of a 200 mL aqueous solution containing 1.45 g of the dissolved salt? The Ka2 value for sulfuric acid is 1.2 \u00c3\u2014 10\u00e2\u20ac\u201c2.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the pH of the 200 mL aqueous solution containing 1.45 g of calcium sulfate dihydrate (CaSO\u2084 \u00b7 2H\u2082O)<\/strong>, we need to follow these steps:<\/p>\n\n\n\n

\n\n\n\n

Step 1: Determine moles of CaSO\u2084 \u00b7 2H\u2082O<\/strong><\/h3>\n\n\n\n

Given:<\/p>\n\n\n\n

    \n
  • Mass = 1.45 g<\/li>\n\n\n\n
  • Molar mass = 172.17 g\/mol<\/li>\n<\/ul>\n\n\n\n

    Moles=1.45 g172.17 g\/mol\u22480.00842 mol\\text{Moles} = \\frac{1.45\\ \\text{g}}{172.17\\ \\text{g\/mol}} \\approx 0.00842\\ \\text{mol}Moles=172.17 g\/mol1.45 g\u200b\u22480.00842 mol<\/p>\n\n\n\n

    \n\n\n\n

    Step 2: Find concentration in 200 mL solution<\/strong><\/h3>\n\n\n\n

    Volume=200 mL=0.200 L\\text{Volume} = 200\\ \\text{mL} = 0.200\\ \\text{L}Volume=200 mL=0.200 LConcentration=0.00842 mol0.200 L=0.0421 M\\text{Concentration} = \\frac{0.00842\\ \\text{mol}}{0.200\\ \\text{L}} = 0.0421\\ \\text{M}Concentration=0.200 L0.00842 mol\u200b=0.0421 M<\/p>\n\n\n\n

    \n\n\n\n

    Step 3: Dissociation in water<\/strong><\/h3>\n\n\n\n

    Calcium sulfate dissolves as:CaSO\u2084\u2192Ca2++SO\u20842\u2212\\text{CaSO\u2084} \\rightarrow \\text{Ca}^{2+} + \\text{SO\u2084}^{2-}CaSO\u2084\u2192Ca2++SO\u20842\u2212<\/p>\n\n\n\n

    The sulfate ion (SO\u2084\u00b2\u207b<\/strong>) is the conjugate base of the hydrogen sulfate ion (HSO\u2084\u207b<\/strong>), which has a Ka\u2082 = 1.2 \u00d7 10\u207b\u00b2<\/strong>.<\/p>\n\n\n\n

    We use this Ka\u2082<\/strong> to assess the basicity of SO\u2084\u00b2\u207b<\/strong> via its Kb<\/strong>:Kw=1.0\u00d710\u221214(at 25\u00b0C)K_w = 1.0 \\times 10^{-14} \\quad \\text{(at 25\u00b0C)}Kw\u200b=1.0\u00d710\u221214(at 25\u00b0C)Kb=KwKa=1.0\u00d710\u2212141.2\u00d710\u22122=8.33\u00d710\u221213K_b = \\frac{K_w}{K_a} = \\frac{1.0 \\times 10^{-14}}{1.2 \\times 10^{-2}} = 8.33 \\times 10^{-13}Kb\u200b=Ka\u200bKw\u200b\u200b=1.2\u00d710\u221221.0\u00d710\u221214\u200b=8.33\u00d710\u221213<\/p>\n\n\n\n

    \n\n\n\n

    Step 4: Use ICE table for SO\u2084\u00b2\u207b hydrolysis<\/strong><\/h3>\n\n\n\n

    SO\u20842\u2212+H2O\u2194HSO4\u2212+OH\u2212\\text{SO\u2084}^{2-} + H\u2082O \\leftrightarrow HSO\u2084\u207b + OH\u207bSO\u20842\u2212+H2\u200bO\u2194HSO4\u2212\u200b+OH\u2212<\/p>\n\n\n\n

    Let x<\/strong> be the concentration of OH\u207b formed:Kb=x20.0421=8.33\u00d710\u221213K_b = \\frac{x^2}{0.0421} = 8.33 \\times 10^{-13}Kb\u200b=0.0421×2\u200b=8.33\u00d710\u221213×2=(8.33\u00d710\u221213)(0.0421)=3.51\u00d710\u221214x^2 = (8.33 \\times 10^{-13})(0.0421) = 3.51 \\times 10^{-14}x2=(8.33\u00d710\u221213)(0.0421)=3.51\u00d710\u221214x=3.51\u00d710\u221214\u22485.93\u00d710\u22128 M OH\u207bx = \\sqrt{3.51 \\times 10^{-14}} \\approx 5.93 \\times 10^{-8}\\ \\text{M OH\u207b}x=3.51\u00d710\u221214\u200b\u22485.93\u00d710\u22128 M OH\u207b<\/p>\n\n\n\n

    \n\n\n\n

    Step 5: Calculate pOH and then pH<\/strong><\/h3>\n\n\n\n

    pOH=\u2212log\u2061[OH\u2212]=\u2212log\u2061(5.93\u00d710\u22128)\u22487.23\\text{pOH} = -\\log[OH\u207b] = -\\log(5.93 \\times 10^{-8}) \\approx 7.23pOH=\u2212log[OH\u2212]=\u2212log(5.93\u00d710\u22128)\u22487.23pH=14\u22127.23=6.77\\text{pH} = 14 – 7.23 = 6.77pH=14\u22127.23=6.77<\/p>\n\n\n\n

    \n\n\n\n

    Final Answer:<\/strong><\/h3>\n\n\n\n

    The pH is approximately 6.77<\/strong><\/p>\n\n\n\n

    \n\n\n\n

    Explanation<\/strong><\/h3>\n\n\n\n

    Calcium sulfate dihydrate (CaSO\u2084 \u00b7 2H\u2082O) is a neutral salt derived from a strong base (Ca(OH)\u2082) and a diprotic strong acid (H\u2082SO\u2084). The second dissociation step of sulfuric acid is weak, and it determines the acidic or basic character of the sulfate ion in water. To calculate the pH, we start by finding how many moles of salt are present. Given a mass of 1.45 g and a molar mass of 172.17 g\/mol, this gives about 0.00842 mol in 200 mL of water, resulting in a 0.0421 M solution.<\/p>\n\n\n\n

    When dissolved, calcium sulfate releases Ca\u00b2\u207a and SO\u2084\u00b2\u207b ions. The calcium ion does not affect pH. The sulfate ion can react with water slightly, forming some hydroxide ions. Since sulfate is the conjugate base of HSO\u2084\u207b, we calculate its basicity using the given Ka\u2082 of sulfuric acid. Using the relationship Kb = Kw \/ Ka, we find that Kb for SO\u2084\u00b2\u207b is 8.33 \u00d7 10\u207b\u00b9\u00b3.<\/p>\n\n\n\n

    We then solve for the hydroxide ion concentration using an ICE table and the equation for Kb. This gives an OH\u207b concentration of approximately 5.93 \u00d7 10\u207b\u2078 M. The pOH is calculated from this value and subtracted from 14 to find the pH.<\/p>\n\n\n\n

    The final result, a pH of 6.77, indicates that the solution is very slightly acidic but very close to neutral. This makes sense for a salt formed from a strong base and a weak acid, where the conjugate base only weakly hydrolyzes in water.<\/p>\n\n\n\n

    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

    Calcium sulfate dihydrate (CaSO4 \u00e2\u20ac\u00a2 2H2O) has a molar mass of 172.17 g\/mol. Considering that calcium sulfate dihydrate is a salt, what would be the pH of a 200 mL aqueous solution containing 1.45 g of the dissolved salt? The Ka2 value for sulfuric acid is 1.2 \u00c3\u2014 10\u00e2\u20ac\u201c2. 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-41492","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41492","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=41492"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41492\/revisions"}],"predecessor-version":[{"id":41494,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/41492\/revisions\/41494"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=41492"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=41492"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=41492"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}