st","extracted_subject":"CHEMISTRY","extracted_chapter":"IONIC EQUILIBRIUM","extracted_topic":"SOLUBILITY EQUILIBRIA OF SPARINGLY SOLUBLE SALTS","questionPreviewUrl":"https://cdn.pureessence.tech/question-previews/packed_canvases/canvas_1153.png?top_left_x=532&top_left_y=0&width=532&height=246&app=solveeit"},"seo_metadata":{"keywords":[]},"doubt_solver_data":{"concept_summary":[]},"embeddings":[],"is_hidden":false,"_id":"66d23e69425d76b7ab317859","slug":"the-ksubspsub-of-mgohsub2sub-is-1-10sup12sup-001-m","content":"The K_sp of Mg(OH)₂ is 1 × 10^–12. 0.01 M Mg(OH)₂ will precipitate at the limiting pH –","options":[{"_id":"68a399947f15af4716bb390b","text":"3","sequence":0,"isCorrect":false},{"_id":"68a399947f15af4716bb390c","text":"9","sequence":1,"isCorrect":true},{"_id":"68a399947f15af4716bb390d","text":"5","sequence":2,"isCorrect":false},{"_id":"68a399947f15af4716bb390e","text":"8","sequence":3,"isCorrect":false}],"correct_answer":"9","explanation":"[Mg²⁺] [OH^–]² = 10^–12\n\n0.01 × [OH^–]² = 10^–12\n\n[OH^–] = 10^–5 M\n\n\\\\ [H⁺] = 10^–9 M and pH = – log [10^–9] = 9","question_type":"single_choice","appeared_in":[],"created_at":"2024-08-30T21:49:29.762Z","__v":0,"vector_store_id":"f327d189-8ed2-4327-9e4d-2eae2b96a76e","updated_at":"2024-08-30T21:49:29.762Z","isStarred":false},"params":{"questionSlug":"the-ksubspsub-of-mgohsub2sub-is-1-10sup12sup-001-m"}}]}]]

Question: The K<sub>sp</sub> of Mg(OH)<sub>2</sub> is 1 × 10<sup>–12</sup>. 0.01 M Mg(OH)<sub>2</sub> will pre...

Solution