Question

6) At 25 C, the K, for formic acid (HCO2H) is 1.8 x 10-4. What is the pH of a 0.10 M aqueous solution of lithium formate (LiHCO2)? 7) What is the pH (aq., 25 °C) of this solution: adding 1.64 grams of sodium acetate to give 200.0 ml solution at 25.0°C? The K, at 25.0°C for acetic acid is 1.8 x 10-5.

Answer 1

oll Linco, aqueous solution ionizes as t Linco3 + H₂O - LOH + H2CO3 I 0,1OM C ta E 0:10- >C Kaxko = 10-14 .: K > 10-14 0.10-X Solution has Lion (strong bose) kg = 1,8*10-4xKg = 10-14 (0.10->) kb= 1.8 X10-4 5.56X16"= x² = 5.56x10-1 Neglecting xi a = 5:56x16" xo) ac? = SIS6 X10-12 .: >= 2.36 x 10-6 is [kon]= [om-) = x2= 2,36x10-6 m pon - log [on- -log (2.36x108) = 5.63 .. pm = 14 -pom = 14-5.63 = 8.37 ma of moles of ll Molar Mass of Сн,соо sodium acetate linen Moss Molar Mass 1,64 82.03 g/mol = 0.02 mol COO Na = 12x2+1GX2+23+3 = 82.03 g mal?

1.8x105 kg = 10-14 Since, kaxkg = 10-14 10-14 Kb = 118 x 10-5 = 0.556 X10-9 = 5.56x10-10 +H₂O Naon CH3COo Na CM2COOH + I 0.02 M +L + L > x E 0.02-> "kg = [chacoom] [naom] [ch₂coona] 5:56X100 = 2.8L O'02-> 2 Neglecting at SI56 *10-10 x0,02= > i ?= 0.1112x10-10 11.12 x 10-12 3. 33 x 10-6 :. [Naom]= [OH-] = x = 3.33 X 106m .: pone - log con ) = -log (3.33X165 x² = 548 ..PM = = 14 - por =14- Se 48 = 18.52