Question

Determine the pH?

1)   0.18M KCHO2

2)   0.15M CH3NH3I

3)   0.19M KI

thanks

Answer 1

1) KCHO2 is called potassium formate. Of course either Ka or Kb value is needed.
HCOOH ? H+ + HCOO-, pKa = 3.744
In solution the following hydrolysis takes place:
HCOO- + H2O ? HCOOH + OH-, Kw/Ka . . . . . . . . (1)
where Kw/Ka is to be proved the equilibrium constant:
We have at equilibrium:
[HCOOH]*[OH-]/[HCOO-] = ([HCOOH]/[HCOO-] [H+])*([OH-]*[H+]) = Kw/Ka

Since HCOOH and OH- are generated in pairs in the hydrolysis (1), we have: [HCOOH] = [OH-]. According to (1):
Kw/Ka = [HCOOH]*[OH-]/[HCOO-]
= [OH-]^2/0.18
Hence: [OH-] = ?(0.18*Kw/Ka)
= ?(0.18*10^-14/10^-3.744)
= 3.16x10^-6 (M)
pH = 14+log([OH-]) = 14+log(3.16x10^-6) = 8.50

3) KI is the salt of hydroiodic acid..a strong acid.Because HI is a strong acid its conjugate base I^-1 is a very weak base and cannot hydrolyze in water ( it cannot compete with OH^-1 for protons so it cannot strip a proton from water to produce a basic solution ) The pH should be 7

2) pH= -log[0.18]
pH= 0.74