Answer:
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
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