Question

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Student Name section # Student ID# Date 1. A student was asked to prepare three solutions and to calculate their pH: Solution A: 0.24 M of NH3 and 0.20 M of NHACI Solution B: adding 0.005 mol of NaOH to 0.50 L of solution A Solution C: adding 0.030 mol of HCl to 0.50 L of solution A a. Calculate the pH of these solutions. (NHA"; K, -5.6x101) b. What property for buffered solutions can you conclude from the results of question a? Give an explanation. S.Carbonats buffers are important in regulating the pH of blood at 7.40. If the carbonic acid concentration in a sample of blood is 0.0012 M, determine the bicarbonate ion concentration required to buffer the pH of blood at pH-7.40. H.CO, (aq) + HCO3(aq) + H (20) Ka-4.3 * 107

Answer 1

1. (1) (A) Are H The pH im, is calcuated wing is calculated wing Henderson equation Her ! PH PK Abt es ? 10-14 C: Now ; Nhz + H20 À NHMT + Öll K: ENHA loita [NHG Lohithth ku . [120] [H20] [H+] :: kg = 10040.18x10-4 18010-5 -> PK b: 4.75 Kom Hut 56x 10-16 5.6 0.24 L w following :: PH475+ log 620) = 4.67. (B) on os no moles of NHud: 0.20x0.s moles a 0.1 mola In ois Li moles of NH3 0.2uxois moles: 0.12 moles I hus; when 0.005 mol of NaOH is added, following reaction occurs :. Nhut + Ho Ž NH3 + H₂O 0.10 0.005 e 1003) ce blog cotes 0.125 INHOU] = 0.095 moles; [NH32 = (0-12 + 0.005) et moles .PH. 4.95 + long ( 10.095) : Kepales) 4:63. (hand) 10 when 0.030 mol of Hel are added; following reaction occurs NH₃ + HU N HUTU 0.12 0.030 - [ NH, 1. | 4:12 - 9 •0 30) - 4 - 0 1 0 1 01 ; TH Hu*1 ( •0 30 + 4+t) - 0 -130 1.PH : KDO 495 + log 160.60) - 4.91 My ) from the results a we find buffer solutions resist change in pH upon adding bare laud 2. Kas 4.3 X 1024 : pka: 6.39 PH : pkat log [ Hwg] - [ H₂ W3 or, 9.40 6.37 + log [H.03] , log [#1 1003 [HW] : 0.0012 10 105 0.013. (An Hope it helps. H 0.0012