[{"data":1,"prerenderedAt":181},["ShallowReactive",2],{"topic-page:\u002Fhigh-school\u002Fbasic-chemistry\u002Facid-base-and-redox\u002Fneutralization-quantitative-relations":3},{"topic":4,"category":145,"seo":150,"breadcrumbs":155,"prerequisites":162,"nextTopics":173,"relatedTopics":180},{"id":5,"slug":6,"title":7,"summary":8,"canonicalPath":9,"categoryPath":10,"difficulty":11,"targetAudience":12,"estimatedMinutes":13,"generatedAt":14,"publishedAt":14,"updatedAt":14,"reviewStatus":15,"tags":16,"expectedKnowledge":22,"learningGoal":25,"blocks":26},"high-school-basic-chemistry-acid-base-and-redox-neutralization-quantitative-relations","neutralization-quantitative-relations","中和の量的関係","酸が出すH⁺と塩基が出すOH⁻の物質量を、価数・モル濃度・L単位の体積から求めて中和条件を判断します。","\u002Fhigh-school\u002Fbasic-chemistry\u002Facid-base-and-redox\u002Fneutralization-quantitative-relations","\u002Fhigh-school\u002Fbasic-chemistry\u002Facid-base-and-redox","beginner","student_high",6,"2026-07-28","ai_generated",[17,18,19,20,21],"化学基礎","中和","価数","モル濃度","当量",[23,20,24],"酸・塩基の価数","正味の中和式","酸・塩基の価数とcVを用いてH⁺量とOH⁻量を比較し、過不足またはちょうど中和する量を求められるようになる。",[27,33,43,64,70,80,99,105,111,127,136],{"id":28,"type":29,"title":30,"subtitle":31,"shortDefinition":32},"hero","HeroBlock","H⁺のmolとOH⁻のmolをそろえる","価数を掛けてから比較する","中和が量的にちょうど進む条件は、酸が出すH⁺の物質量と塩基が出すOH⁻の物質量が等しいことです。多価の酸・塩基では価数を掛けます。",{"id":34,"type":35,"term":36,"definition":37,"plainExplanation":38,"keywords":39},"definition","DefinitionBlock","中和の当量関係","酸由来のH⁺量と塩基由来のOH⁻量が等しいとき、化学量論的な当量点になります。","1価のHCl 1 molはH⁺ 1 mol分、2価のH₂SO₄ 1 molは量的にH⁺ 2 mol分です。同様にCa(OH)₂ 1 molはOH⁻ 2 mol分です。濃度と体積だけを比べず、価数×c×VでH⁺またはOH⁻のmolへそろえます。",[40,41,42],"H⁺量=OH⁻量","価数×cV","体積はL",{"id":44,"type":45,"title":46,"lead":47,"formulas":48,"strategy":63},"formula","FormulaBlock","中和条件","酸と塩基の価数をそれぞれ独立に入れます。",[49],{"label":36,"expression":50,"meaning":51,"symbols":52,"tip":62},"a × cₐ × Vₐ = b × cᵦ × Vᵦ","左辺は酸が出すH⁺量、右辺は塩基が出すOH⁻量です。VはL単位です。",[53,56,59],{"symbol":54,"meaning":55},"a","酸の価数",{"symbol":57,"meaning":58},"b","塩基の価数",{"symbol":60,"meaning":61},"c, V","モル濃度（mol\u002FL）と溶液体積（L）","この等式はちょうど中和する当量点について使います。","酸\u002F塩基→価数→濃度→mLからL→積、の順に対応表を作り、両辺の単位がmolになることを確認します。",{"id":65,"type":66,"src":67,"alt":68,"caption":69},"diagram","DiagramBlock","\u002Fassets\u002Fhigh-school\u002Fbasic-chemistry\u002Facid-base-and-redox\u002Fneutralization-quantitative-relations\u002Fneutralization-equivalents.svg","2価の酸と1価の塩基について価数と物質量からHプラス量とOHマイナス量を比較する天秤図","酸側はa×cₐ×Vₐ、塩基側はb×cᵦ×Vᵦとして、H⁺カードとOH⁻カードのmol数を比較します。体積はLへ換算します。",{"id":71,"type":72,"steps":73,"title":79},"steps","StepBlock",[74,75,76,77,78],"酸と塩基の化学式からそれぞれの価数a、bを読む","各溶液のモル濃度cを対応させる","mLで与えられた体積を1000で割ってLへ直す","酸のa cVと塩基のb cVを計算する","等しければ当量、差があればH⁺またはOH⁻の過剰量として判断する","中和量計算の5段階",{"id":81,"type":82,"columns":83,"rows":87,"summary":98},"compare","CompareBlock",[84,85,86],"比較結果","中和後の量的判断","注意",[88,91,95],[40,89,90],"当量点","pH 7とは限らない",[92,93,94],"H⁺量>OH⁻量","酸が量的に過剰","残量を差で求める",[96,97,94],"H⁺量\u003COH⁻量","塩基が量的に過剰","当量点は反応量の等しさです。生成塩や酸・塩基の強弱によって液性は異なり得ます。",{"id":100,"type":101,"scenario":102,"explanation":103,"result":104},"example","ExampleBlock","0.100 mol\u002FL H₂SO₄ 20.0 mLを、NaOH 25.0 mLでちょうど中和した。NaOHのモル濃度を求める。","H₂SO₄は2価、NaOHは1価です。酸側は2×0.100 mol\u002FL×0.0200 L=0.00400 mol分のH⁺です。塩基側を1×c×0.0250 Lとし、0.00400=0.0250cよりc=0.160 mol\u002FLです。","NaOHは0.160 mol\u002FLです。20.0と25.0をmLのまま使わず、価数2も含めました。",{"id":106,"type":107,"warningTitle":108,"message":109,"severity":106,"action":110},"warning","WarningBlock","体積比だけで中和を決めない","同じモル濃度でも、2価の酸は1価の酸の2倍のH⁺量を出し得ます。またmLをLへ直さず式へ入れると、他の量との単位がそろいません。当量点と指示薬の終点も同じ概念ではありません。","酸と塩基の欄を分け、価数・濃度・L体積を1行ずつ対応させてから積を作りましょう。",{"id":112,"type":113,"title":114,"questions":115},"quiz","QuizBlock","確認テスト",[116],{"question":117,"choices":118,"answerIndex":122,"choiceExplanations":123},"0.100 mol\u002FL HCl 20.0 mLをちょうど中和する0.200 mol\u002FL NaOHは何mLですか。",[119,120,121],"10.0 mL","20.0 mL","40.0 mL",0,[124,125,126],"正解です。どちらも1価で、0.100×0.0200=0.200×VよりV=0.0100 Lです。","濃度が2倍のNaOHには、HClと同じ体積は必要ありません。","濃度と必要体積の関係を逆にしています。濃い側の必要体積は小さくなります。",{"id":128,"type":129,"title":130,"points":131},"summary","SummaryBlock","まとめ",[132,133,134,135],"中和量は酸のH⁺量と塩基のOH⁻量で比較する","ちょうど中和ではa cₐVₐ=b cᵦVᵦが成り立つ","多価酸・多価塩基の価数とmLからLへの換算を落とさない","当量点は量的関係であり、必ずpH 7になるとは限らない",{"id":137,"type":138,"title":139,"nextTopics":140,"relatedKeywords":142},"next-action","NextActionBlock","次は滴定操作と終点へ",[141],"high-school-basic-chemistry-acid-base-and-redox-acid-base-titration-procedure",[143,89,144],"中和滴定","終点",{"path":10,"title":146,"description":147,"parentPath":148,"accentToken":149},"酸・塩基と酸化還元","酸・塩基と中和、pH、酸化還元、金属のイオン化傾向や電池の入口を整理していくカテゴリです。","\u002Fhigh-school\u002Fbasic-chemistry","signal-blue",{"title":151,"description":152,"canonicalUrl":153,"ogImage":154},"中和の量的関係 | 化学基礎","酸・塩基の価数、モル濃度、L換算した体積からH⁺量とOH⁻量をそろえて中和量を計算します。","https:\u002F\u002Fzeqnilo.com\u002Fhigh-school\u002Fbasic-chemistry\u002Facid-base-and-redox\u002Fneutralization-quantitative-relations","https:\u002F\u002Fzeqnilo.com\u002Fassets\u002Fsite\u002Fdefault-og.svg",[156,159,160,161],{"title":157,"path":158},"高校","\u002Fhigh-school",{"title":17,"path":148},{"title":146,"path":10},{"title":7,"path":9},[163],{"id":164,"title":165,"summary":166,"canonicalPath":167,"categoryPath":10,"categoryTitle":146,"difficulty":11,"targetAudience":12,"estimatedMinutes":168,"publishedAt":14,"updatedAt":14,"reviewStatus":15,"tags":169},"high-school-basic-chemistry-acid-base-and-redox-neutralization-and-net-ionic-equation","中和と正味のイオン反応式","酸と塩基の中和を、分子式、全イオン式、正味のイオン反応式の三段階で読み、実際に変化する粒子を特定します。","\u002Fhigh-school\u002Fbasic-chemistry\u002Facid-base-and-redox\u002Fneutralization-and-net-ionic-equation",5,[17,18,170,171,172],"イオン反応式","H⁺","OH⁻",[174],{"id":141,"title":175,"summary":176,"canonicalPath":177,"categoryPath":10,"categoryTitle":146,"difficulty":11,"targetAudience":12,"estimatedMinutes":13,"publishedAt":14,"updatedAt":14,"reviewStatus":15,"tags":178},"中和滴定の操作と終点","ホールピペット、ビュレット、三角フラスコを用いる中和滴定の手順と、当量点・終点の違いを整理します。","\u002Fhigh-school\u002Fbasic-chemistry\u002Facid-base-and-redox\u002Facid-base-titration-procedure",[17,143,179,89,144],"ビュレット",[],1785565316174]