Question

Elm in the pH of a buffer solution which contoing 9.60 mole Of ammonium nitrate and 0.4 0 mole of ammonig in a 1.0 d m 3 solution. Given Ra for NH 4 = 5.6x 10-10 Lo Write the acid-base conjugate pairs for the following reaction N H 4 + H2O2 N H + HOÁ (2 marks) nicolculate the pH of 0.75 M. Sodium Qietate Solution CH3COON o. Givin Kaforecetic acid, CH, COOH is 15 x10-5. (3natul iv. A buffer solution is made by mixing 350 cm 3 of 0 tom om thonte N Ha Solution with 450 cm? 0.1 0.15 Mommonium Chlorid NHCP Solution calculate thi PH of this buffer solution Given ko for "NH 3 = 1.8x10-5 (5 miks

Hi.can you help me answer questions b and d.thanks a lot for your help

Answer 1

ama Los 1 dm3 ] H₂O b) ka for NHAT = 5.6*10-10 "Kb for NH₃ = KWlka = 10-14 /5.6810-10 = 1.8 X10-5 .PKb of NH3 = -log (1.8x10-5) = 4.75 ÞH = bkb + log [sall] [from Henderson's equation] [base] Iroon I ÞH = 4.75 + log [0.60/1] = 4.93 [[Salt] = 0.60 mall] [0.401] base = 0.40 moll bH= 4.93 | 11) NH4+ + H2 0 = N H + H3 01 acid base t h e conjugate acid. conjugate base NHAT is Ht donor, hence it is acid and is Ht acceptor, it is base. (1) CH₃COONa (sodium acetate) is a salt of weak acid and strong base. The formula for ÞH calculation of this type of compound is, m PH = 1/2 pkw + + 2 pka + haloge onremos con we know bkw = 14 at 25°C bka = -log ka E-log (1.75 X10-5) = 4.75 - ÞH = x14 + x 4.75+ á xperto log 0.75 = 9.312 (iv) Total volume (350+450) dm3 = 800 dm3 - [NH₂] = 0.401800 mole/dm3 [NHAC] = 0.151800 molel dm 3. Kb = 1.8X10-5, ÞKb = -log (1.8X10-5) = 4.75 ] c. bt = pkb & loa [salt [Воло] - [0.40/800] = 4.32 a ci) NH3 + H₂O Ž NHAT & OH base acid - conjugate conjugate acid base NH3 is base because ito accepts de proton, Ho is acid because it donates proton =4.75 + log [0.151800]

11) The concentration of Ht of HF is calculated as [ht] = √ka.c = 16.6 X 1024 x0.2 -132 x 10-4 (M) 2.DH = -log (1-32 X10-4) - 3.88 as we know, ÞH + bol = 14 - boh - 14-3.88 = 10:12 2. - log [OH] =10.12 - [ou] = 10-10.12 = 7.5 X10-" (M) (nl) Molecular weight of HON = 1 + 12+14 - 279. Number of moles = 0.763127 = 0.03 mole. - concentration = (0.031.5) molel lit = 0.2 motellit [total volume = 150 mL = 0.15 lit] - [ht] = √kaic 16.2x10-10x0.2 = 1.11X10-5 ÞH = -log (1.01 X10-5) tog (11x0? What =4.95