Question

What is the pOH of 0.50 molar H3BO3?

Given: pKb = 4.77

A.
10.25
B.
1.96
C.
2.54
D.
11.46

Answer 1

D. 11.46

Step-by-step explanation:

Got it right on plato

Answer 2

A. 10.25

Step-by-step 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*10^(-10)= \frac {x^2}{(0.50M-x)}`

(-x is neglected) so we get

`5.88*10^(-10)*0.50=x^2\\\\x^2=2.94*10^(-10)`

`x=√(x^2)=1.72*10^(-5) M=H^3 O^(+)`

`pH=-log[H^3 O^+]\\\\pH=-log[1.72*10^(-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