Question

What is the pH of 0.0800 M Fe(NO3)3 (Ka of Fe3+ = 3.00x10-3)? Express your answer to two decimal places.

Answer 1

pH is defined as the -log[H+]

Fe(NO​​​​​3​)3 will hydrolyse in water to give Fe(OH)2 and H​​​​​​3​​​​​O​+ this we can find out from equation of dissociation constant

That is

K​​​​​​a = (H3​​​​​O​+) (Fe(OH)2)/ (Fe​​​​​3+)

Given concentration of ferric ion is 0.08

At time t ferric ions are 0.08-x

And ferrous hydroxide and H+ are x

Put in equation of dissocuation constant that is

3×10-3= x​​​​​​2​​​/(0.0800-x)

Solving quadratic equation we get x= 0.00662

pH= -(log 0.00662)

pH=2.18