[{"data":1,"prerenderedAt":126},["ShallowReactive",2],{"cheatsheet-solution-stoichiometry-and-titrations":3},{"sheet":4},{"sheetSlug":5,"topicSlugs":6,"primaryTopic":7,"title":8,"subtitle":9,"sections":10},"solution-stoichiometry-and-titrations",[5],11,"Solution Stoichiometry & Titrations","n = MV stoichiometry, acid-base and redox titrations, standards, purity, and gravimetric analysis.",[11,23,31,41,49,57,66,74,96,104,117],{"heading":12,"kind":13,"items":14},"Balance first","rules",[15],{"ref":16,"mode":19,"provenance":20,"text":21,"source_hash":22},{"type":17,"topic":7,"id":18,"field":17},"rule","SS-R001","transform","owned_workbook","A titration is stoichiometry in solution. Write and balance the reaction before any calculation.","e88ef2282c52a96d1d0ceb8ffc8b5dc22d3d0230007b8a7f406dd1b566fe5973",{"heading":24,"kind":13,"items":25},"n = MV",[26],{"ref":27,"mode":19,"provenance":20,"text":29,"source_hash":30},{"type":17,"topic":7,"id":28,"field":17},"SS-R002","Moles from a solution: n = M × V (V in litres). Rearrange: V = n / M. Use it wherever a mass-based path would use molar mass.","95b3faf719ffeb0ce07faf6fa88f0a3fad3517265ef87141feca4b9b1f9a12f2",{"heading":32,"kind":33,"items":34},"Titration vocabulary","data",[35],{"ref":36,"mode":19,"provenance":38,"text":39,"source_hash":40},{"type":17,"topic":7,"id":37,"field":17},"SS-R019","original","Titrant: known solution from the buret. Analyte: unknown being measured. Equivalence point: titrant volume delivering the exact stoichiometric amount (equal moles only for a 1:1 reaction). Endpoint: measured volume (indicator colour change). Primary standard: high-purity weighable solid (e.g. KHP).","741665fec3320c3bc3dedb88f6abb718734258c6f2a8203f89515cbd36beeb18",{"heading":42,"kind":13,"items":43},"KHP acid standard",[44],{"ref":45,"mode":19,"provenance":20,"text":47,"source_hash":48},{"type":17,"topic":7,"id":46,"field":17},"SS-R003","KHP (KHC₈H₄O₄, 204.22 g/mol) is a monoprotic solid acid that reacts 1:1 with NaOH; used to standardize base solutions.","79dc2159fff1118128b7a063bc28660746afffbcae551335382179b40fcee569",{"heading":50,"kind":13,"items":51},"Polyprotic neutralization",[52],{"ref":53,"mode":19,"provenance":20,"text":55,"source_hash":56},{"type":17,"topic":7,"id":54,"field":17},"SS-R005","Polyprotic acids neutralize stepwise (a separate equivalence point per proton). At complete neutralization: H₃PO₄ needs 3 NaOH; H₂SO₄ and H₂C₂O₄ each need 2 NaOH.","ed83864f4a135a592762b3cbfb377619e5b3f666f3b20b71aead3504cfa20fed",{"heading":58,"kind":59,"items":60},"Unknown concentration","steps",[61],{"ref":62,"mode":19,"provenance":20,"text":64,"source_hash":65},{"type":17,"topic":7,"id":63,"field":17},"SS-R006","mol titrant (M × V) → mol analyte (stoichiometric factor) → divide by analyte volume: M = n / V.","d3e5a4b3c7480e590bf2329bf7838ee712d96eb9f20f462637b757f4326e84c3",{"heading":67,"kind":59,"items":68},"Purity of a solid",[69],{"ref":70,"mode":19,"provenance":20,"text":72,"source_hash":73},{"type":17,"topic":7,"id":71,"field":17},"SS-R007","Titration → mass of pure analyte; percent purity = (mass analyte / total sample mass) × 100.","acfd0c686ab7d8d14ac73a0d54e0eac0f9cbfbc2786279ebac7e726d1a587dc9",{"heading":75,"kind":33,"columns":76,"items":79},"Redox titrations",[77,78],"Titrant","Balanced half-context (acidic)",[80,88],{"ref":81,"mode":19,"provenance":20,"rows":83,"source_hash":87},{"type":17,"topic":7,"id":82,"field":17},"SS-R009",[84],[85,86],"K₂Cr₂O₇","Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺","4ae44a0fcf8ca7a04ec27708eb1d743b72badde560b5c4627e2935b6b7302b0f",{"ref":89,"mode":19,"provenance":20,"rows":91,"source_hash":95},{"type":17,"topic":7,"id":90,"field":17},"SS-R010",[92],[93,94],"KMnO₄","MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O","9a7e3c075b5bf97aa5c075f15d334254fcfb9a57b7045098cdfb754e5e4618e6",{"heading":97,"kind":13,"items":98},"Limiting reactant in solution",[99],{"ref":100,"mode":19,"provenance":20,"text":102,"source_hash":103},{"type":17,"topic":7,"id":101,"field":17},"SS-R012","Same as with pure substances, but moles come from n = MV. Compare each reactant's moles against the balanced ratio to find the limiting one.","aad7a2b6fe154800e779db796f7d872bac689b0de86e8e678a27f7de0ea390f0",{"heading":105,"kind":59,"items":106},"Precipitation & gravimetric",[107,112],{"ref":108,"mode":19,"provenance":20,"text":110,"source_hash":111},{"type":17,"topic":7,"id":109,"field":17},"SS-R013","Identify the precipitate (solubility rules), find moles of each reactant, pick the limiting one, then compute precipitate mass.","d3da85e5225fe2609d8817d5a3f7afd728fb6038ff063aae6481d92022b4b531",{"ref":113,"mode":19,"provenance":38,"text":115,"source_hash":116},{"type":17,"topic":7,"id":114,"field":17},"SS-R016","Gravimetric: mass of precipitate → mol precipitate (÷ MM) → mol analyte (ratio) → mass analyte; mass % = (mass analyte / sample mass) × 100.","34012e7a94f91abe4b0eebb20b975f062ad181dec8513ec2b5f2b7bb46a875dc",{"heading":118,"kind":119,"items":120},"Hydrated reagents","pitfalls",[121],{"ref":122,"mode":19,"provenance":20,"text":124,"source_hash":125},{"type":17,"topic":7,"id":123,"field":17},"SS-R004","For a hydrated base (e.g. LiOH·H₂O), molar mass includes the water, but the stoichiometry uses the base formula, so one mole of hydrate gives one mole of LiOH.","acd1b19da7789160167cbae937f2baec36e7b5e45330895bb6b047dcb002dd5a",1787246033567]