Question

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Name: Lab Section: Date: Experiment VII: Buffers Review Questions (23 pts) 1. (3 pts). A solution is prepared by mixing 25.0 mL 0.45 M NaCH,coo and 30.0 mL 0.38 M CH,COOH is prepared. Calculate the pH of this solution if the K, 1.76 x 10 for acetic acid. 2. (3 pts)How many grams of sodium lactate, NaCsHsOs, should be added to 1.00 L of 0.150 M lactic acid, HCsHsOs, to form a buffer solution with a pH of 3.90? (K, 1.4x 10 for lactic acid) 3. (4 pts) What is the pH of the a buffer solution containing 0.10 moles of acetic acid and 0.12 moles of sodium acetate in 45 mL after adding 0.030 moles of KOH?

Answer 1

1. [azcooNa = 25mLy 0.45m (25+30) - 0.205m [Choom - 30 mL x 0.38m (25+30) me - 0.207 m ww. hiver Ka= 1.76X10-5 apka =-wg ka =-wg (1176X10-9) =4.751 nors from Henderson egle, CC1hCOUNT) pH = pka + log 112000H]

pH = 4.75 +90.1045 0 0.2073 - 4.74. - pH of the sole is 4.74 (2.) from hendersm ea, pka + log sau Balt) mg 3.90 =-wg (1.4x104) + log o 150 (salt) = 1691 0166808929 M e 1000 ml solnconturn 0.166808929 mol of Nagtso Moler mom of Nacz to 03 = (13+36 +5 +48)g mitt = 1129 mohol -0.166808924 mil of Nachfolg ere = 0.166808929 mgffx 112 g mati of Naczty og ~ 18.68 g of Naghsh 1868 gt Naghs should be added. (3) 0.60 miles of chroon + d. 22 miles of choo na. Tout 0.030 mile of Koth in added, think it react with Choot to form the salt scook. Honce, total o no of mole of ch coolt remam =(0.10 -0.03) = 0.07 miles. liter ton of moles A salt ie fil coonat ck cook) in 0:12+0.03) = 0.15 moles non taud) = _0.07 mily 45 ml. - de hos m 0.07 (:'M = md/1) 48X103 m

Balt) = 0.15 moles 458103002 = 0.11 os mille = entot M SpH = pka + log and =- wgkat wg Isalt) 01 45*10-3 0.07 45x103 = - Log (1.76x107) + wg oil - 5.09