Question

6. What is the pH of a buffered system made by dissolving 17.42 g of KH2PO4 and 20.41 g of K2HPO4 in water to give a volume of 200.0 mL? The Ka2 for dihydrogen phosphate is 6.2 x 10-8 and the equilibrium reaction of interest is H2PO4 (4) + H20 <-> H30* ( HPO4 (4) 7. Determine the pH of a 0.188 M NH3 solution at 25°C. The Kh of NH3 is 1.76 x 10-5. 8. Calculate the pH of a buffer that is 0.225 M HC2H302 and 0.162 M KC2H302. The Ka for HC2H302 is 1.8 x 10-5. 9. Determine the K, of an acid whose 0.294 M solution has a pH of 2.80

Answer 1

Ans © pH of buffer solution - ? ; vol. of solution - 200 ml Amount of Kha Pop = 14.42g = 0.2L Amount of RHPOL - 20.419 concentration of Kitap Oy = Griven mars Molar Masu x V [KHAPO.] = 17.429 = 17.42 M " OQL 27.2 = 0.64 M Concentration of High Poy – dolygra AM 174 g moll X 0.2L 34.8" [KHPO4) = 0.58M Kag for dihydrogen phosphate (k!PO4) = 6-2x10-8 Hapou (aq) + H20 (1) ² H30+ (aq) + Ipo caq.) According to Henderson - Hasselbalch equation - pH = play + log (Coryugate bare] kg = 6.2x108 [ Acid ] pkay = -logkas =-10966-2xloe) = -log 6•2+8 = 7.21 + lagl6.90625) = 7.21+ (-0.04) = 7.21+ log on the = -0.79+8 - 7-21 PH = 7.17

s® Phi o 188 M NH, Aolution = ? kg = 1.76 X10's Dissociation of NH3 is written as- NH, CH2+ ,o(2P NH" aq) + oi (4) Initial 0-188 M conc. ta +34 At qull. 0.188-X conc ko = [MF] Cov) Co.188-X+0.188) 1.76 *105 = 22 2x = 1.76 x 105 x 0.188 212 033088x10-5 x= √3.3088 X164 x = 1,818102 M [OH) = 1.81x10-2 [H+] [or'] = 1x1014 M [12+] = 1810-14 M = 0.55 X10-12 M pl1 = -1091H+] - -- 10966.55X20-12) = -log 0.55 +12 = -0.25 +12 1.81 xlo = 11.75 pH = 11.75

Ans 6 pH of buffer solution = ? [46 H30,] = 0.225 M [ KG303) = 0.162 M Ka for HCG H2O = 1.8x10-5 pka = -log ka = -log (1-8810-5) = -log 18 - log(105) -0.25 +5 Spka = 4.75] According to Henderson - Hasselbalch equation – pH = pka + log [salt] TAcid J 5 4.75 + long 0162 0.225 = 4.75 + log 0072 = 4.75 – 0.14 PH = 461 pH of buffer solution is 4.61

Ani acid 6 Determine ka for an pH = 2.80 pH = -log[ht] [Mt) - lo-PM = 1o-2.80 [H+) = 1.58 x1637 HA+100 F H (mm) + A Caq) 0.294M Initia conc. 1 0:294-2 + equll. 51158x163 conc. Ke = [ht] [A] [HA] 22 (0.994->) (0.294-460-227) ka = (1.58 x103) 0.294 = 2.4964 X10-6 0.994 I ka = 8.49x10-6 7 So, the ka value for acid whose 0.294M pH 2.80 i's 8649x10-5 solution has