Question

At 25°C,the dissociation constant of a base, BOH, is 1.0 x 10^-12. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be:

• A. 1.0 x 10^-6 mol L^-1
• B. 1.0 x 10^-5 mol L^-1
• C. 1.0 x 10^-8 mol L^-1
• D. 1.0 x 10^-7 mol L^-1

Answer 1

The Correct Option is D

To find the concentration of hydroxyl ions in a 0.01 M aqueous solution of the base, BOH, we need to relate the dissociation constant (\(K_b\)) to the concentration of OH^- ions produced during the dissociation.

The dissociation reaction of BOH is given by:

BOH \rightleftharpoons B^+ + OH^−

Given that the dissociation constant of BOH is \(K_b = 1.0 \times 10^{-12}\).

Let \([OH^-]\) be the concentration of hydroxyl ions. Initially, the concentration of BOH is 0.01 M, and it dissociates to produce \([OH^-]\) and \([B^+]\) in a 1:1 ratio. At equilibrium, the concentrations can be expressed as:

• [BOH] = 0.01 - [OH^-]
• [OH^-] = x
• [B^+] = x

Since \(x\) will be very small compared to 0.01, we can approximate \([BOH] \approx 0.01\).

The expression for the dissociation constant \(K_b\) is:

K_b = \frac{[B^+][OH^-]}{[BOH]}

Substituting the known values, we have:

1.0 \times 10^{-12} = \frac{x \times x}{0.01}

Solving for \(x\), the equation becomes:

x^2 = 1.0 \times 10^{-12} \times 0.01

x^2 = 1.0 \times 10^{-14}

x = \sqrt{1.0 \times 10^{-14}}

x = 1.0 \times 10^{-7} \text{ mol L}^{-1}

Therefore, the concentration of hydroxyl ions in the solution is \(1.0 \times 10^{-7} \text{ mol L}^{-1}\).

Thus, the correct answer is: 1.0 x 10^-7 mol L^-1.