Question

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

Answer 1

`pH=11`

Step-by-step 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)=(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!