Question

calculate the pH of a 0.15 M solution of FeSO 4

Answer 1

Start by writing a dissociation equation for FeSO 4

FeSO 4 -----> Fe2+ + SO 4 2-

SO 42- + HOH -----> HSO 4 - + OH-

Kw=Ka x Kb
Ka = 1*10^3
Kw=1.0x10^-14

(1.0x10^-14)/(1.0x10^3) = Kb = 1.0x10^-17



SO 42- initial: .15____ change: -x_____ final: .15-x
H SO 4- initial: 0_____ change: +x____ final: x
OH- initial: 0____ change:+x_____ final: x

Now, we can plug the "final" values from the table into the Kb equation.

Kb=[ HSO 4-][OH-]/[ SO 42-]
1.0x10^-17=[x][x]/[0.15-x]


1.0x10^-17=[x][x]/0.15
1.5x10^-18=x^2
x=1.22x10^-9

x=1.22x10^-9=[OH-] ... Now that you know this, we can solve for the pOH...

-log(1.22x10^-9)=8.92=pOH

pH + pOH = 14.00, so...
14.00-8.92=pH=5.08