<p> Chem 178 SI February 27, 2008 1. Which of the following are strong bases? - + - - NaOH ClO NH4 CaO Sr(OH)2 HSO4 NO2</p><p>2. When is the assumption that x<< [HA] or x<<[B] valid? When the percent ionization is less than 5%.</p><p>-7 -11 3. A solution contains 0.3 M of H2CO3. Ka1 = 4.3x10 Ka2=5.6x10 . What is the pH of the solution? Ka1 > Ka2 by a factor of 103 so the pH will depend only on the first protonization. + - Ka= [H ][ HCO3 ]/[H2CO3] + - + - [H2CO3] Ka=[H ][ HCO3 ] [H ]=[ HCO3 ] Assume x<<[H2CO3] 2 -7 -7 [H2CO3] Ka=x = (0.3)( 4.3x10 ) = 1.3x10 x=3.6x10-4 = [H+] Check to make sure the assumption is valid. + -4 %ion. = 100 x [H ]/[H2CO3] = 100(3.6x10 )/(0.3) = 0.11% The Assumption is valid so pH = -log 3.6x10-4 = 3.44</p><p>- -8 4. 0.35 moles of ClO are in 1 L of solution. Kb = 6.7x10 a) Write the balanced chemical equation of the reaction. - - ClO + H2O  OH + HClO b) What is the pH? - - - Kb= [OH ][ HClO]/[ ClO ] assume x<<[ ClO ] 2 Kb= x /0.35 (0.35)( 6.7x10-8)= x2 x=1.5x10-4 pOH=-log1.5x10-4=3.81 pH=10.19</p><p>-8 5. Ka for H2CO3 is 3.2x10 . Write the balanced chemical equation and find Kb for the reverse reaction.</p><p>- + H2CO3 + H2O  HCO3 + H3O Ka x Kb = Kw -4 -8 -7 Kb= 1.0x10 /3.2x10 =3.1x10</p><p>6. Kb for COOH- is 2.5x10-5. The initial concentration of HCOOH is 0.67M. Write the balanced chemical equation and find the pH. - - COOH + H2O  HCOOH + OH Ka x Kb = Kw -4 -5 -10 Ka = 1.0x10 /2.5x10 =4.0x10 + - Ka= [H ][ COOH ]/[HCOOH] assume x<<[HCOOH] 2 -10 ([HCOOH]) (Ka)= x = 2.68x10 X = 1.6x10-5 pH=4.79</p>