Question

nicotine is resposible for the addictive properties of tabacoo assume Kb1= 1.05x10^-2 and Kb2= 1.32x10^-11. Determine the pH of a 0.00125M solution

Answer 1

nicotine + H2O ------> nicotine+ + OH^-

using ICE table

  nicotine (C10H14N2)+ H2O ------> nicotine+( (C10H15N2)⁺   + OH^-

initial   0.00125M 0 0       

equilibrium     0.00125 - x x x

Ka1= [ nicotine⁺ ] [ OH^-] /[  nicotine] = x² / 0.00125 - x   

1.05x10^-2 =  x² / 0.00125 - x

1.31 *10^{-5 -} 1.05*10^-2 x = x²

x²+ 1.05* 10^-2 x - 1.31 *10^-5 = 0

( using quadratic equation we find value of x )

x = -b +_ (b²- 4ac )/2a

x = -1.05* 10^-2 +_ (1.05* 10^-2 )²- 4(1) - 1.31 *10^-5 /2a

x =-1.05* 10^-2 +_ 1.1025 * 10^-4+ 5.24 *10^-5 /2(1)

x =-1.05* 10^-2 +_ 1.1025 * 10^-4+ 0.524 *10^-4 /2(1)

x =-1.05* 10^-2 +_ 1.6265 * 10^-4 /2(1)

= -1.05* 10^-2 + 1.27 * 10^-2 /2

= 0.22 * 10^-2 /2 = 0.11 * 10^-2 M

as concentration of  nicotine+( (C10H15N2)⁺ = OH- =  0.11 * 10^-2 M

we calculate x for Kb2

  nicotine+( (C10H15N2)⁺ + H2O -------> nicotine2+( (C10H16N2)²⁺ +OH ^-

initial 0.11 * 10^-2 M 0    + 0.11 * 10^-2 M

equilibrium     0.11 * 10^-2 -x x + 0.11 * 10^-2 +x

Kb= [ 0.11 * 10^-2 M +x] [x] /    0.11 * 10^-2 M -x   

1.32x10^-11 = 0.11 * 10^-2 x + x² /  0.11 * 10^-2 -x   

0.11 * 10^-2-x ( 1.32x10^-11) =  0.11 * 10^-2x+x²

0.1452 * 10^-13 -  1.32 * 10^-11x =   0.11 * 10^-2x+x²  

x² + 0.11 * 10^-2x +   0.1452 * 10^-13= 0

      ( using quadratic equation we find value of x )

x = -b +_ (b²- 4ac )/2a

as in above c is 0.1452 * 10^-13 which is very small( we neglect it ) therefore in above expression c is neglected and as a = 1 only b remains and as we can see -b +_ (b²⁾/2 = 0

just calculate you will get x =0

hence we use x =  OH- =  0.11 * 10^-2 M   

pOH=-log  0.11 * 10^-2 =2.95

as pH= 14 -pOH = 14 - 2.95 = 11.05