Question

- 4 A chemistry graduate student is given 500. mL of a 0.90 M trimethylamine (o N) solution. Trimethylamine is a weak base with K,=7.4 10 4. What mass of (CH3), NHBr should the student dissolve in the (CH3), N solution to turn it into a buffer with pH = 10.45? You may assume that the volume of the solution doesn't change when the (CH) NHBr is dissolved in it. Be sure your answer has a unit symbol, and round it to 2 significant digits.

Answer 1

= 2C mol 140.02 Let x g of (CH3)3 NH Br - should be added. molar mayol (CH3.) 3 NH Br: 140.02 gimol mole ol (CH3).3 NHBr = mass = a - molar mas 140.02 glmol wolume of solution : 500 ml = 0.5 L ( 1000 m I-IL) concentration (M mole Volume (2) [ CcHals NHB ] = mol = U M 140.02 X 0.5L 70.1 [(CH3)N] = 0.90 M Kb = 7.4810-4 bkb - 109 kb = - 109 (7.4 x 104) = 3,13 pH = 10.45 pOH = 14-bH pot = 14 - 10.45 = 3.55 Using Henderson Hayalbalch equation þOH = pka + 109 [(CH3)3 NHBr] [CH3) 3N] 3.55 = 3.13 + log ( x ) 10.90 M 10.1 0.42 lo 0.4 10 2.63 70.1X0.90 = 2.63 X 70.1 X 0.90 = 165.99 z x = 1.7X1029 To 2 sropificant - Answer 170g = 1.7 X 109g figures