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![Day HA is Week mono protic acid. HA + H₂O A & Hot at 0.36(M) 750 .x at 0.36-2. equilibrium, [A-] [Hot] x² Ka of neida 0.36.x](http://img.homeworklib.com/questions/68ddee30-0a2c-11eb-b7b6-138812b0de72.png?x-oss-process=image/resize,w_560)
Day HA is Week mono protic acid.
HA + H₂O A & Hot at 0.36(M) 750 .x at 0.36-2. equilibrium, [A-] [Hot] x² Ka of neida 0.36.x [HA] as x is too smal, x² 0:36 x² = 0.36 % ka 1 (M) 0.36 X 3.4 X10 +3 11063 X 10 () TH30t] = -1063x103 (4) Log (1-1063x10 -: pH - log (14307] - = 12:956 Like above question, for a monoprotic and 2.62 Now, THz ort] = x = J Ka x [HA] pit = log [H3ot) = * TH204) = 24398x102(M)
2 Ka КА pH = 14- РОН Given, (1) а Оде Д. ОТЈа (м) - 3 2.398 хүр 4 хв. 01/2 4:18 x 10 of acid 19іях) 1 In briffer solution concentration of base Св) = 0.14 (9) Concentration ute acid (g) CBнт ) sport=pkot Гр 4:17 = 14- 4-9= | 9-р). and, of Conjugat =o.
45 (4) ") | 1 - logee of log (6.95 -- log (3.4x15) tong Lot 0:40 О») 4
d) Following reaction oruurs: f etti NH₂4 CH₂NH₂ of Hel is titrated against 2 As, 0129 (M) CH₃NH2 0.29 (M) CH?Nit2 same of Hel required to reach egnivalence point, : [enz. Why :] = 0.2g (M) 50.12(M) pH = 7 & Pront log Trait] = 7+ 2 log kor & log (9.12) = 7+1 loge (3 x 109) Ilog (0:12). 5.809 pH = 15.81
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