st","extracted_subject":"CHEMISTRY","extracted_chapter":"IONIC EQUILIBRIUM","extracted_topic":"IONIZATION OF ACIDS AND BASES ","questionPreviewUrl":"https://cdn.pureessence.tech/question-previews/packed_canvases/canvas_1138.png?top_left_x=532&top_left_y=0&width=532&height=264&app=solveeit"},"seo_metadata":{"keywords":[]},"doubt_solver_data":{"concept_summary":[]},"embeddings":[],"is_hidden":false,"_id":"66d23e69425d76b7ab317605","slug":"the-ph-of-05-m-bacnsub2sub-solution-is--given-pksu","content":"The pH of 0.5 (M) Ba(CN)₂ solution is-\n\nGiven pK_b of CN^– = 9.3","options":[{"_id":"68aab4f37f15af4716dbcbd6","text":"8.35","sequence":0,"isCorrect":false},{"_id":"68aab4f37f15af4716dbcbd7","text":"3.35","sequence":1,"isCorrect":false},{"_id":"68aab4f37f15af4716dbcbd8","text":"9.35","sequence":2,"isCorrect":true},{"_id":"68aab4f37f15af4716dbcbd9","text":"9.5","sequence":3,"isCorrect":false}],"correct_answer":"9.35","explanation":"Ba²⁺ is not hydrolysed , CN^– = 1 (M)\n\nCN^– + H₂O \\(\\rightleftarrows\\)HCN + OH^–\n\npH = \\(\\frac{1}{2}\\) (pK_w + pK_a + log c)\n\npH = \\(\\frac{1}{2}\\) (pK_w + pK_a) since c = 1(M)\n\npH = \\(\\frac{1}{2}\\) (2pK_w – pK_b) = \\(\\frac{1}{2}\\) (28 – 9.3)\n\npH = \\(\\frac{1}{2}\\) (28 – 9.3) = 9.35","question_type":"single_choice","appeared_in":[],"created_at":"2024-08-30T21:49:29.548Z","__v":0,"vector_store_id":"76c9717a-c7fa-49cf-990f-711eb599749c","updated_at":"2024-08-30T21:49:29.548Z","isStarred":false},"params":{"questionSlug":"the-ph-of-05-m-bacnsub2sub-solution-is--given-pksu"}}]}]]

Question: The pH of 0.5 (M) Ba(CN)<sub>2</sub> solution is- Given pK<sub>b</sub> of CN<sup>–</sup> = 9.3...

Solution