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
Answer:
2.35 × 10⁻⁹
Explanation:
Isoquinoline is a weak base according to the following equation.
C₉H₇N + H₂O ⇄ C₉H₇NH⁺ + OH⁻
The pH is 9.468. The pOH is:
pH + pOH = 14
pOH = 14 - pH = 14 - 9.468 = 4.532
The concentration of OH⁻ is:
pOH = -log [OH⁻]
[OH⁻] = antilog -pOH = antilog -4.532
[OH⁻] = 2.938 × 10⁻⁵ M
The concentration of the base (Cb) is 0.367 M. We can find the base dissociation constant (Kb) using the following expression.
[OH⁻] = √(Kb × Cb)
Kb = [OH⁻]²/Cb
Kb = (2.938 × 10⁻⁵)²/0.367
Kb = 2.35 × 10⁻⁹
