Question

answer is 0.2869. pls show work

COH + NaOH- COK Potassium hydrogen phthalate FM242 CON + HO COK The NaOH was then used to standardize H2SO4 H2SO4 + 2NaOH - Na SO, + H2O (a) Titration of 0824 g of potassium hydrogen phthalat required 38,314 g of NaOH solution to reach the end point detected by phenolphthalein indicator. Find the concentraron of NaOH expressed as mol NaOHkg solution 0) A 10063-gram aliquot of H, SO, solution required 57.9013 of NaOH solution to reach the phenolphthalein end point. F the concentration of H,SO, in mol/kg of solution

Answer 1

The answer is-

botassium a) According Anthalate I According to the reaction between potansi torto phthalate and NovoH; Imol 4 KHP requires I mol of NooH. so 0.824 mol reup 0.82ng of KHP = 204.22² 4.0 3 x 10 mol KHP so 4,03 x 10 mol UHP = 4:03x103 mol Noola on 2 (4.03x10 x no ) ga Nooh 0.161 gm Novo H So 38.314 g solution has 0.161 gm North 30 1000 g solution has a 0.261x1000 am Naolt 38-314 0.161X1000 mol NaOH 38.314XLp 2 Orlos mol Noolt nie So concentrations o.1os mol/ug of solution areal 2N00H + H250 = Na, sou + 2H,0 30 1 mol of H, so requires a mol of Norot 57.01g of Noch nolution has 0.16 1x9 2914 57.911 g of Nash solution has a 0.161857.001 20.2434 g Noot = 60084x10 mol Novote 30 H₂Sou required & 8x 6,0 84 x 103) mo 1 2 3,042 x 10 mo 1 = 0.298 ga So concentration 4 Hy sly? 10.023x09 mollege 0.302 mol/ing 38.314 g Nach