Can somebody me with this question, i'm very confused! i thought i did it right because i found kb and the conc. of oh- ions and then used poh to find ph but the answer i get is very !

what is the ph of a 0.51 mol l-1 ch3nh3+cl- aqueous solution? pkb(ch3nh2) = 3.44 answer to 2 decimal places.
(answer = 5.43)

Let's investigate the substances involved in the reaction first. The compound CH3NH3+Cl- is a salt from the weak base CH3NH2 and the strong acid HCl. When this salt is hydrated with water, it will dissociate into CH3NH2Cl and H3O+:

CH3NH3+Cl- + H2O ⇒ CH3NH2Cl + H3O+

Nest, let's apply the ICE(Initial-Change-Equilibrium) table where x is denoted as the number of moles used up in the reaction:

                 CH3NH3+Cl- + H2O ⇒ CH3NH2Cl + H3O+
Initial                  0.51                             0               0
Change                 -x                             +x             +x

Equilibrium        0.51 - x                         x               x

Then, let's find the equilibrium constant of the reaction. Since the reaction is hydrolysis we use KH, which is the ratio of Kw to Ka or Kb. Kw is the equilibrium constant for water hydrolysis which is equal to 1×10⁻¹⁴. Since the salt comes from the weak base, we use Kb. Since pKb = 3.44, then. 3.44 = -log(Kb). Thus, Kb = 3.6307×10⁻⁴ 

KH = Kw/Kb = (x)(x)/(0.51 - x)
1×10⁻¹⁴/ 3.6307×10⁻⁴ = x²/(0.51-x)
x = 3.748×10⁻⁶

Since x from the ICE table is equal to the equilibrium concentration of H+, we can find the pH of the aqueous solution:

pH = -log(H+) = -log(x)
pH = -log ( 3.748×10⁻⁶)
pH = 5.43

Do you have any options?