Question

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.

Answer 1

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

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