{"common":{"save":0,"post_id":"8849","level":3,"total":10,"point":10,"point_extra":0},"segment":[{"id":"6571","post_id":"8849","mon_id":"1159892","chapter_id":"1159893","question":"<p data-pm-slice=\"0 0 []\">Tr\u1ed9n 100 ml dung d\u1ecbch HCl có pH = 1 v\u1edbi 100 ml dung d\u1ecbch g\u1ed3m KOH 0,1M và NaOH aM, thu \u0111\u01b0\u1ee3c 200 ml dung d\u1ecbch có pH = 12. Giá tr\u1ecb c\u1ee7a a là<\/p>","options":["A. 0,12","B. 0.08.","C.0,02.","D. 0,10"],"correct":"3","level":"3","hint":"","answer":"<p>\u0110áp án \u0111úng: <strong>C.0,02.<\/strong><\/p>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-04-17 11:09:46","option_type":"txt","len":0},{"id":"6572","post_id":"8849","mon_id":"1159892","chapter_id":"1159893","question":"<p data-pm-slice=\"0 0 []\">Ch\u1ea5t \u0111i\u1ec7n li y\u1ebfu bao g\u1ed3m<\/p>","options":["A. Acid y\u1ebfu","B. Base y\u1ebfu","C. Mu\u1ed1i không tan","D. C\u1ea3 A và B \u0111\u1ec1u \u0111úng"],"correct":"4","level":"3","hint":"","answer":"<p>\u0110áp án \u0111úng: <strong>D. C\u1ea3 A và B \u0111\u1ec1u \u0111úng<\/strong><\/p><p><strong data-end=\"86\" data-start=\"66\">Ch\u1ea5t \u0111i\u1ec7n li y\u1ebfu<\/strong> là nh\u1eefng ch\u1ea5t không phân li hoàn toàn thành ion trong dung d\u1ecbch. C\u1ea3 <strong data-end=\"167\" data-start=\"155\">acid y\u1ebfu<\/strong> và <strong data-end=\"183\" data-start=\"171\">base y\u1ebfu<\/strong> \u0111\u1ec1u là nh\u1eefng ch\u1ea5t \u0111i\u1ec7n li y\u1ebfu, vì khi tan trong n\u01b0\u1edbc, chúng ch\u1ec9 phân li m\u1ed9t ph\u1ea7n thành ion.<\/p><p><strong data-end=\"293\" data-start=\"281\">Acid y\u1ebfu<\/strong> nh\u01b0 <strong data-end=\"307\" data-start=\"298\">H\u2083CO\u2083<\/strong> (axit cacbonic) ho\u1eb7c <strong data-end=\"340\" data-start=\"329\">CH\u2083COOH<\/strong> (axit axetic) ch\u1ec9 phân li m\u1ed9t ph\u1ea7n trong n\u01b0\u1edbc.<\/p><p><strong data-end=\"404\" data-start=\"392\">Base y\u1ebfu<\/strong> nh\u01b0 <strong data-end=\"416\" data-start=\"409\">NH\u2083<\/strong> (amoniac) c\u0169ng ch\u1ec9 phân li m\u1ed9t ph\u1ea7n trong n\u01b0\u1edbc.<\/p><p><strong data-end=\"486\" data-start=\"468\">Mu\u1ed1i không tan<\/strong> (nh\u01b0 <strong data-end=\"501\" data-start=\"492\">BaSO\u2084<\/strong>) là ch\u1ea5t không phân li hoàn toàn trong n\u01b0\u1edbc, tuy nhiên, chúng không ph\u1ea3i là ch\u1ea5t \u0111i\u1ec7n li y\u1ebfu mà là nh\u1eefng ch\u1ea5t không tan.<\/p>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-04-17 11:13:55","option_type":"txt","len":2},{"id":"6573","post_id":"8849","mon_id":"1159892","chapter_id":"1159893","question":"<p data-pm-slice=\"0 0 []\">pH c\u1ee7a h\u1ed7n h\u1ee3p dung d\u1ecbch HCl 0,005M và H2SO4 0,0025M là<\/p>","options":["A. 2","B. 3","C. 4","D. 12"],"correct":"1","level":"3","hint":"","answer":"<p>\u0110áp án \u0111úng: <strong>A. 2<\/strong><\/p><p data-pm-slice=\"0 0 []\">HCl --> H+ + Cl-<\/p><p>-> [H+] = 0,005 M<\/p><p>H2SO4 --> 2H+ + SO4 2-<\/p><p>-> [H+] = 0,0025x2 = 0,005 M<\/p><p>=> T\u1ed5ng [H+] = 0,01 M<\/p><p>=> pH= -log(0,01) = 2<\/p>","type":"choose","extra_type":"classic","user_id":"139","test":"0","date":"2025-04-17 11:15:21","option_type":"txt","len":0}]}
