Question

Preparing Buffer Solutions: Calculating the Number of Grams of Conjugate Base Needed Use the table of K values given with this problem to choose the best weak acid to start from for making a buffer that holds the pH of the solution at 8.50. Make your selection so that you maximize the capacity of the buffer. Weak Acid K 1.8 X 10-5 6.5 X 10-5 1.5 X 10-5 You select a 800.0 ml volumetric flask to which you add 8.00 mL of a 4.70 M solution of the weak acid you just selected. To finish preparing your buffer you must now add the sodium salt of the conjugate base. How many grams of the sodium salt of your conjugate base must you add so that when you finally fill the flask to the mark with deionized water the buffer will have a pH of 8.50. 1.8 X 10-4 CH3COOH (Acetic Acid) C6H5COOH (Benzoic Acid) CH3CH2CH2COOH (Butanoic Acid) HCOOH (Formic Acid) HBrO (Hypobromous Acid) HNO2 (Nitrous Acid) HCIO (Hypochlorous Acid) CH3CH2COOH (Propanoic Acid) 2.8 X 10-9 4.6 X 10-4 2.9 X 10-8 1.3 X 10-5 4.9 X 10-10 HCN (Hydrocyanic Acid) grams

Answer 1

Weak auld (HA) 1. Intin Final Molanity M, = 470m M₂=2 Volume V 80m) VE 8'OOM MINI = Mala Final concentsation My = Mini - V M2F 4.70m x 8:00) 800M) Me= 0.0479 Hy Pobro rows auld (HH BYO). ka = 2.8x109 pha= logika pha = = log [2.8x10] = 8055. Given pH = 8.50 According to Henderson's Egnation pt of Wear bases buttes solutions pH = pha tlog Caride pite fiso. pha & gossi I salt Edcide] = 0.0479 [salt] Coneatsation = x

pH s pka trog (sa le Joy Chaet) = get_pkan. log (2) - log CacidaJ = pt-ptka log(x) log Colour] = 8.50 -8oss. 1092 +10327. = 0.05. 108x = 0.05 - 10327 logrs - 1.279 I logaca - 1277 x= 10 1 2 = 109 250.0528M I concentsation of Canongated base Chabro) = 0·0528171 - M & Weight x 1000 Moliwt M = 0.05L8M1 Weig het af Moleculs weight of NaBro = 118.gg r. Volume = Ve 800m) Weight = MXMolWtXV 0:05288118-89 XBoomt 1000 = 5o2grams. 1910) Wint MAMOWA 7 - Tooont sodium Hybromite Weight [5002 / fram,