Question

Calculate the pH of a 0.187 M solution of a base that has a Kb = 5.86x10^-6

Answer 1

Answer:

$pH=11$

Explanation:

Hello there!

In this case, since the ionization of the described base is:

$base+H_2O\rightarrow OH^-+baseH^+$

Thus, the equilibrium expression in terms of the reaction extent is:

$Kb=5.86x10^{-6}=\frac{x^2}{0.187-x}$

Thus, by solving for x we obtain:

$5.86x10^{-6}({0.187-x})=x^2\\\\1.10x10^{-6}-5.86x10^{-6}x-x^2=0$

So the values of x are:

$x_1=0.0010M\\x_2=-0.0010M$

So the feasible answer is 0.0010 M, thus we compute the pOH:

$pOH=-log(0.0010M)=3$

And therefore the pH:

$pH=14-pOH=14-3\\\\pH=11$

Best regards!