Question

- Chem. 132 Homework Problems Fall 2019 PH POH= 14 Chemistry: Homework Problems: You Need to Master Chapter 17 Acids and Bases 1) What is the conjugate acid of HCO3"? 2) What is the conjugate base of H2PO4"? 3) What is the concentration of hydroxide ions in pure water at 30.0°C, if kw at this temperature is 1.47 x 10-14? 4) What is the pH of pure water at 40.0°C if the Kw at this temperature is 2.92 x 10-14? What is the Kw of pure water at 50.0°C, if the pH is 6.630? 6) Calculate the concentration of H30* in a solution that contains 5.5 x 10-5 MOHⓇ at 25°C. Identify the solution as acidic, basic, or neutral. 7) Calculate the concentration of OH" in a solution that contains 3.9 x 10-4 M H30 at 25°C. Identify the solution as acidic, basic, or neutral.

Answer 1

* (1) Ans: According to Broonstad - Lowry theory of acid. base, the conjugate acid of Heo will be H₂CO₃. HCO3 + Ht - H₂CO3 As HCO2 takes a proton to froom H₂CO₃ do, H₂CO3 is a conjugate acid of Hcoz and Hog, H₂ CO 3 are called conjugate acid-base pain. (2) Ans. According to Bronsted - Locony theory of acid base, a species that is fromed after selease of a proton is called the conjugate base. Hence the Conjugate base of H₂pog will be Hoog- H₂PO4 HPO TH+. (1) Nole: Conjugate aeld is formed when a species takes . a procton mesulting a conjugate aeld. llen

1 (3) Anu! At 300 kw 1.47x10-14 (given) Autoronization equilibrium of H₂O; H₂O + H₂O + H₂otton- - Kg. [Hz ot / Lost - Kis dissociation Constant [H2OT of H2O. Kw = [Hot] [ost] kwa BK d[H207 In pure waters, - Ionic product of Ho [Hot] - [ost Kwa [OH-] on [ost - Skw - 11.47x10-14 = 1.21 x 10-7 (M) Thes, the concentration of hydroxide ions in pare H2o at 30 c is 121x 10-7 (M) .. (1) Ans: At 40c Knə 2.92*10-14 Foroo pame waters [Hot] [ort] i Kw = [H3072 •pkwaloga on, - log kw 2-2log [Hot] - log (2.92x10-14) on Pkw 2-2log (Hot) - 13.73 or - log [ltzot] al pkw on pH > Y Pkw a 1 x 13 53 26.765

later (o 161 Ans! (5) Ans! at 50.0° PH2 6.630 foot pace wate foro parce walen pH = 1 pkr on Pkw, 2pH = 2x6:630 2 13:26 on – log Kw2 13:26 Kwa 10-13126 = 5:495x10-14 • At 250 the range of pH scale is Otolt. - PH + port = 14 [on]5.5x10-511) - pools-log(5:5310-8) - 4.2596 ..pH = 14-port 29.7404 . at 25°c the pH of neutral solution will be B [Hz ot] - Cont] => pH oport >7. pero o In this case the pH of the Solution is 9.7404. Hence it lies game above the neutral point so the solution will be alkaline in nature, PH₃9.7404 = -log (H₂01 ] ₂9.7104 => [H₂ ot] - 10-9.7404 - 1.8180 X 10-10 (M)

Therefore the concentration of Hotina solution Containing 6-5*10-(21) CH- of 25° is 1:818x10. • The solution is basic in nature. (7) Ans: [Hoot) - 3.9x10-4 (M) at are pH = -log FH301] = -log(39x10-4) = 3.4089 post 2 14-pH a 14 -3.4089 10.5911 on-loglost- 10.5911 . Cost- a to -10.5911 a 2.56389x10-11 Since the pH of this solution is below tle, the neutral pH so the natwee of the solution is acidic.