Question

Please Solve Question 1 and 3 Practice Shot -CHEM 212 250 m o d.150 M N O Hatut bedo chieve the isted with 100 M HC, Cate the volume linge point (The kao HOC is 753 by Calculate the pH of the NIOCI Solution bore any acidised ci Calculate the pH of the solution after 10.0 mL of 0.100 M HClisad Cotthop of the solution after 250 ml of 0.400 M HCl is added e Calculate the pH of the solution for 500 ml of 0 100M H isade 1) the pH of the solution at the one point a With the data point thatch the titration curve 3) Bood is buffered mainly by the HCO, -H.00, wstom K.H.CO. - 44X10'). The normal ph of toodis 7.40 What is the ratio [H.COMHOOT What does the pH become if 10% of the HOO, lons are converted to H.CO? c) What does the ph become il 10% of the HCO micul are converted to HCO) P H .CO.isadi protic acid, which implies that its conjugue boe, HCO (the bicarbonate on my love d for all of these out on you my wethepara HUD ( L) HH" ( + ) * (1) Its acidity is very with respect to Garbonic acid

Answer 1

0.15M Naocl 25 ml x 0 15 mol = 3.75 mmol of Naoci Titrated with O. IMHCI Reaction taking flace; Naoci + Hal Hoci + Nach [na • 1 mol HCl react with 1 mol Naoci in 3.75 mmol of Naod will react with 3.75 mmol of Hy for equivalence Point Volume Hal= moles Hol molarity Hal = 3.75 Xto 3 mol Ool mol L- Via = 37.5 mL (1:12=103 m2) @ IT or 37.5x10 3L Before any acid is added We have; 25mL of 0.15M Naoel Since Naoce is salt of Strong base (Naok) * of weak And (HOCH) .: pH = 7+ { pka + log ] C: Koncentration of salt = 0.13M

pH = 7+1 1.63 + log(0.1s)] (pka of Holl = 7.53) | PH = 10.35) © After lomL of o. UMHCI lomyy oil mot = Immol HCl Immol of HCl will react with Immal of Naoci to form 1 mmol of Ност. Hoci = Immol l Hoci] = HOLI = 1 mmol total Volume (25+10mL Nucci Hai [Naoci)= = 0.0286M n Naoc = 3.75-1 mond Vtoral (15+ to m -0.0786M e pH = pka + log [salt] Naoce [Acid) Hoa 0.0286 pH = 7.53 +log (0:0786) pH = 7.97 After 25mL of 0.1 MHA @ 25myxor1 mol = 2.5 mmol HCl Choci] = 2.5 mmol = 0.05m (25+25) mL [Nao ci] = (3.75-2,5)mme) 20.025M (25+25) mL PH="7153+ log (0,025) = 7.23 (PM= 7:23) This pH was found to be 7.23