Respuesta :
Answer:
A. 10.25
Explanation:
Pkb =4.77
So pka = 14 - pka = 9.23
[tex]Ka =10^{-pka}[/tex]
[tex]H_3 BO_3 (aq)+ H_2 O(l) <>H_2 BO_3^- (aq)+H_3 O^+ (aq)[/tex]
Initial 0.50M 0 0
Change -x +x +x
Equilibrium 0.50M-x +x +x
[tex]Ka =\frac {((x)(x))}{(0.50M-x)}[/tex]
[tex]5.88\times10^{-10}= \frac {x^2}{(0.50M-x)}[/tex]
(-x is neglected) so we get
[tex]5.88\times10^{-10}\times0.50=x^2\\\\x^2=2.94\times10^{-10}[/tex]
[tex]x=\sqrt{x^2}=1.72\times10^{-5} M=H^3 O^{+}[/tex]
[tex]pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76[/tex]
pOH = 14 - pH
= 14 - 4.76
pOH = 9.24 is the answer
Option A - 10.25 is the answer which is close to 9.24
