In the laboratory, a general chemistry student measured the pH of a 0.367 M aqueous solution of isoquinoline, C9H7N to be 9.468. Use the information she obtained to determine the Kb for this base.

Explanation:

Isoquinoline is a weak base.

We would follow the steps below to solve for Kb;

1) Write the ionizaton equation for the base. Using B as the symbol for the weak base.

B + H2O ⇌ HB+  +  OH¯

2) Writing out the equilibrium expression:

Kb = ( [HB+] [OH¯] ) / [B]

3) Solving for the three concentrations above.

     a) We will use the pH to calculate the [OH¯]. We know pH = -log [H+], therefore [H+] = 10^¯pH

From the question; pH = 9.468

   [H+] = 10¯9.468 = 3.404 x 10¯10 M

Knowing the [H+] allows us to get the [OH¯].

To do this, we use Kw = [H+] [OH¯], so we have this:

       1.00 x 10¯14 = (3.404 x 10¯10) (x)

       x = (1.00 x 10¯14) / (3.404 x 10¯10)

       x = [OH¯] = 2.934 x 10¯5 M

   b) From the ionization equation, we know there is a 1:1 molar ratio between [HB+] and [OH¯]. Therefore:

       [HB+] = 2.934 x 10¯5 M

   c) the final value, [B] is given in the problem. Concentration of B = 0.367

       Kb = [(2.934 x 10¯5) (2.934 x 10¯5)] / 0.367

       Kb = 2.346 x 10-9

hope you got the answer

pv=nrt

explanation: