Question

Quinine occurs naturally in the bark of the cinchona tree. For centuries, it was the only treatment for malaria. Quinine contains two weakly basic nitrogen atoms, with Kb1=3.31 x 10^-6 and Kb2=.00000000135 at 25 C. Calculate the pH of a .011 M solution of quinine in water.

Answer 1

Study the ionic equation representing its ionization in water of 0.011M quinine solution in following two steps,

Step-1 quinine + water ---------------> quinineH⁺   + OH^-

Initially 0.011 excess 0 0   

At equilibria 0.011 - x x x

where x represents the deegree of ionization

therefore Kb1 = 3.31 x 10^-6   = x²  / (0.011 - x )

x²   = ( 3.31 x 10^-6  ) (0.011 - x )

Let the value of x on solving the above equation be represented by '' a '.

However, for solving the above quadratic equation, certain mathematical simplification or considerations are helpful-

Since Kb1 is fairly small we neglect the (-x ) term from the denominator thus getting -

x^2 = 3.31 x 10^ -6 x 0.011

= 0.03641 x 10^ -6

& x = 0.1908 x10 ^ -3

Solve this quadratic equation to get the value of x. This gives the concentration of quiniineH⁺ which further ionises according to following step or reaction

Step - 2

quinineH⁺   + water ----------------> quinineH₂²⁺   + OH^-   

at equiibria in this case ( a - y ) excess y y

therefore Kb 2 = 1.35 x 10^-9   = y²  / (  a - y )   

again, to calculate the value of y in this case we can easily ignore '' y '' in the denominator, since Kb 2 is so small and also the OH^-  obtained from the hydrolysis will be negligible in comparision to Kb 1

hence, y²   = 1.35 x 10^-9   

and y = say '' b''

Now therefore the total OH^-  concentration in solution = ( a + b )

Note:- Even though ( a+b ) gives the total OH^-1 concentration in the solution ,but again we should look at the given Kb 2 value which is still much smaller than the Kb 1 value. Hence, it can well be inferred that the additional [OH^-1] created from this reaction is negligible and can be ignored.So, the negligible value of '' b " is ignored for further calculations .

Thus, [ OH ^-1] = x = 0.1908 x 10^-3

Calculate p(OH) = using equation p(OH) = - log [ OH^- ]

ie = - log ( a+b ) or, = -log a ( when b s ignored )

= -log (0.1908 x 10^ -3)

= say '' c'' = 3.72

and hence pH = ( 14 - c ) or, = (14-3.72)

= 10.28

Thus the above steps would lead you to find out pH of ,011M quinine solution and to work out the desired answer.