Question

What is the pOH of 0.50 molar H3BO3?

Answer 1

Answer:

A. 10.25

Explanation:

Pkb =4.77

So pka = 14 - pka = 9.23

$Ka =10^{-pka}$

$H_3 BO_3 (aq)+ H_2 O(l) H_2 BO_3^- (aq)+H_3 O^+ (aq)$

Initial                0.50M                                 0                                 0

Change                  -x                                 +x                               +x

Equilibrium    0.50M-x                               +x                               +x

$Ka =\frac {((x)(x))}{(0.50M-x)}$

$5.88\times10^{-10}= \frac {x^2}{(0.50M-x)}$

(-x is neglected) so we get

$5.88\times10^{-10}\times0.50=x^2\\\\x^2=2.94\times10^{-10}$

$x=\sqrt{x^2}=1.72\times10^{-5} M=H^3 O^{+}$

$pH=-log[H^3 O^+]\\\\pH=-log[1.72\times10^{-5}]\\\\pH=4.76$

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