Question

2. The pOH of an acetic solution is 11.45. Calculate hydrogen and hydroxide concentrations. 3. A solution was prepared by dissolving 1.20 g of NH4Cl in enough water to form 500. mL of solution. What is the pH of this solution? 4. A solution was prepared by dissolving 1.625 g of KF in enough water to form 750.0 mL of solution. What is the pH of this solution?

Answer 1

conic product of water kw product of molar concentration of Ht ions and onions Kw = [H+][OH-] 4 Kw = 1410-14 at 298k Kw = [Hi+] [OH-] = 10-14 pH = -log[ * ] , pouze logolon-] .. pH + por=pKw = 14 0-2 CH3COOH + H20 -> CH3C00- + H307/17 poh = 11.45 given : - loq. (043) z 11.45 [OH-] = 3.548133210-12 M

ianie product of water Kw = [4+][04-] 10-12 [+1 13.548133x10-12 ] [H+] 2 2081838364610-3 M [on-] = 3.54 8133X10-12m T4+] = 2-81838364 X10-3 M are enoue ulti (3) Lince Nhill is fully dissociated in water because enough water taken NHYU+ H2O NHy on the We know dissociation of Hill и , ) — н* + и - CheeJ= [ht] =[ce ] (due to equal 1 . stiochiomete

conc" of Why U in water = no. of males of which volume of rolution (2) Grinen mans of NHU Cl = 1. 2g Molar man of NHull a 14+4 + 35.5 5 53.5g/miel (NH4 11: VN na Ginen mass (NHue) Molao mass (Nhy Ce) 1.2 53.5- very welume of relution = soom2 0.5 L Ž 1.2 mul [Nhy Ce] = $3.5 = 0.049 8598 M (mol/L) 0.5L Since stiochiometry of Ny Cl (aq) scartaion is same Imol of Nity u will foroduce. Imal of Nhyol 240 Hence [Nhy CeJ= [Nilyon)- []

M [HU] = 0.04485 98.1 Cnce ] = [ + ]-[e] [ht] = [nel = 0.04885981 m .bha - dog, (CH43) - - dogrel 0-04885981) V10 = 1311048 pH of this solution is 10311048 Strongly acidie (4) KF + H2O → KOHT HE/ complete T i l dissociation) [RF] = [kon] = [HF ] concentrations are equal due to stachiometry HE Ht to dissociation of HF [HF] = [H +)-[F-1 . Melasmass of KF = 58-0967 g/mol

concentration (KF) = mo. of mules of Ke (m) volume of nolution (2) (N n = Griven man Molarman fork - mol 1.625 58.0967 vz 750 ml = 0.75L ĽKF] = 4.625 4.625 58.0967 - 0.75 0.0312941M (mall 2) conch (KF) = cone " ( HF) wonen (H+). 14+] = 0.0372941 m ÞHz - logi & Ht) = 1.428359 pH of this solution = 1.428359 Lummary . This freoblem based upon ionic waste product of water and the complete dissociations concept and pH and pon of different solutions