Calculate pH of 0.10mol/L FeSO4(s) in Distilled Water

[jeahomgrajan]

How do i find the pH of FeSO4(s), knowing that its concentration is 0.10mol/L
it is a salt.. and it was placed in distilled water..FeSO4---->< Fe+2 + SO4-2

SO4-2 is the conjugate base of a strong acid H2SO4- therefore SO4-2 has no effect on the pH

in this case what is Fe+2,would be an acid and the Ka is equla to 1.8*10^-7
then i would combine Fe with h2o
and then what will the equation be?

 

[Borek]

SO₄^2- will hydrolize slightly - giving HSO₄^-, raising pH.

Fe(OH)₂ is weekly soluble, so it may lower pH.

Finding exact answer may be a little bit tricky.

 

[Blueshift5]

The equation would be:
Fe+2 + H2O ⇌ FeOH+ + H+
The equilibrium constant, Ka, would be equal to 1.8*10^-7. To calculate the pH, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([FeOH+]/[Fe+2])
Since the concentration of FeSO4 is 0.10 mol/L, the concentration of Fe+2 would also be 0.10 mol/L. The concentration of FeOH+ can be calculated using the equilibrium constant:
Ka = [FeOH+][H+]/[Fe+2]
Rearranging the equation, we get:
[FeOH+] = Ka * [Fe+2] / [H+]
Substituting the values, we get:
[FeOH+] = (1.8*10^-7) * (0.10) / [H+]
Since the concentration of H+ is equal to the concentration of FeOH+, we can substitute this value into the Henderson-Hasselbalch equation:
pH = pKa + log([FeOH+]/[Fe+2])
pH = -log(1.8*10^-7) + log([FeOH+]/[Fe+2])
pH = 6.74 + log(1)
pH = 6.74
Therefore, the pH of 0.10 mol/L FeSO4 in distilled water would be approximately 6.74.