Question

At 25°c, the dissociation constant of a base, boh, is 10^ -12. The concentration of hydroxyl ipns in 0.01 m of aqueous solution would be

Answer 1

Answer: The concentration of $OH^-$ is $9.99\times 10^{-8}M$

Explanation: Dissociation of base can be written as:

               $BOH\rightleftharpoons B^++OH^-$

at $t=0$     0.01       0       0

at $t=t_{eq}$  (0.01-x)    x       x

Dissociation constant, $K_b$ for the following reaction is written as:

$K_b=\frac{[B^+][OH^-]}{[BOH]}=10^{-12}$

Putting the concentration of each substance at equilibrium in the above equation, we get

$10^{-12}\frac{x.x}{(0.01-x)}$

Rearranging the terms, we get

$X^2+10^{-12}x-10^{-14}=0$

On solving this quadratic equation, we get

$x=9.99\times 10^{-8},-1.00\times 10^{-7}$

We neglect the negative value of 'x' as it is concentration and it cannot be negative. Therefore,

$x=9.99\times 10^-8$

And, $x=[OH^-]=9.99\times 10^{-8}M$

Answer 2

Answer:

1 × 10^7 M.

Explanation:

Refer attachment.

Note: The values of concentration written below the equation is the concentration at equilibrium.

Hope it helps!!